Adv. Inequal. Appl. 2018, 2018:14 https://doi.org/10.28919/aia/3776 ISSN: 2050-7461

ON MULTIPLICATIVE ZAGREB INDICES OF TWO NEW OPERATIONS OF GRAPHS

A. BHARALI

Department of Mathematics, Dibrugarh University, Assam 786004, India

Copyright c2018 A. Bharali. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. Recently, Wang et al. (2017) introduced two new operations of graphs. In this paper the upper bounds of the multiplicative Zagreb indices of the two newly proposed operations of graphs are derived.

Keywords:multiplicative Zagreb indices; graph operations.

2010 AMS Subject Classification:05C05, 05C90, 05C07.

### 1. Introduction

Throughout the paper we consider only simple finite graphs.V(G)andE(G)are respectively

the set of vertices and set of edges of a graph G. The degree of a vertexu∈V(G)is denoted

byd_{G}(u), if their is no confusion we simply write it asd(u). Two vertices uand vare called

adjacent if there is an edge connecting them. The connecting edge is usually denoted by uv.

Any unexplained graph theoretic symbols and definitions may be found in [16].

E-mail address: a.bharali@dibru.ac.in Received June 24, 2017

Topological indices are the numerical values which are associated with a graph structure.

These graph invariants are utilized for modeling information of molecules in structural

chem-istry and biology. Over the years many topological indices are proposed and studied based on

degree, distance and other parameters of graph. Some of them may be found in [5, 7].

Histori-cally Zagreb indices can be considered as the first degree-based topological indices, which came

into picture during the study of total π-electron energy of alternant hydrocarbons by Gutman

andTrinajstic´in 1972 [9]. Since these indices were coined, various studies related to different

aspects of these indices are reported, for detail see the papers [4, 6, 8, 12, 17] and the references

therein.

The first and second Zagreb indices of a graphGare defined as

M1(G) =

_{∑}

u∈V(G)
d_{G}2(u) =

_{∑}

uv∈E(G)
(d_{G}(u) +d_{G}(v)),

M2(G) =

### ∑

uv∈E(G)d_{G}(u)d_{G}(v).

The multiplicative versions of these Zagreb indices are proposed byTodeschini et al.,in [14]

and latter named as “Multiplicative Zagreb indices” by Gutman [10]. These indices can be

defined as follows.

### ∏

1

(G) =

_{∏}

u∈V(G)
d_{G}2(u)

### ∏

2

(G) =

_{∏}

uv∈E(G)
d_{G}(u)d_{G}(v).

Graph operations play a very important role in chemical graph theory, as some chemically

interesting graphs can be obtained by different graph operations on some general or particular

graphs. Like many other topological indices, studies related to Multiplicative Zagreb indices

and coindices of various graph operations [1, 2, 3, 13].

In this paper we consider two new operations of graphs proposed by Wang et al. in [15]

and derive the upper bounds on the multiplicative Zagreb indices of the two newly proposed

operations of graphs. The rest of the paper is organized as follows. In section 2 we reproduce

### 2. Two new graph operations and some preliminaries

In this section we first reproduce the definitions of the two newly defined operations of graphs

[15] and then some standard results are included.

The Cartesian product of graphs G and H, denoted by G_{}H, is the graph with vertex set

V(G)_{}V(H)={(a,v):a∈V(G),v∈V(H)}, and (a,v) is adjacent to (b,w) whenever a=b

and (v,w)∈E(H), or v=w and (a,b)∈E(G). More detail on Cartesian product and some

other operations of graphs may be found in [11]. In 2017, Wang et al., proposed the following

two operations of graphs and also studied their adjacency spectrum. We reproduce the figure in

[15] to make the discussion self expository.

FIGURE 1. Two new operations

Definition 2.1. Let G_{1i} =G_{1} and G_{2i}=G_{2} (1≤i≤k) be k copies of graphs G_{1} and G_{2},
respectively,Gj(j=3,4)is an arbitrary graph.

• The second operation(G_{4}G_{1})_{k}(G_{3}G_{2})ofG_{1},G_{2},G_{3}andG_{4} is obtained by
mak-ing the Cartesian product of two graphsG_{3}andG_{2}, produceskcopiesG_{2i}(1≤i≤k)of
G2 and making the Cartesian product of two graphsG4 andG1, produces kcopiesG1i
(1≤i≤k)ofG_{1}, then makeskjoinsG_{1i}∨G_{2i},i=1,2, . . . ,k.

Now we propose the following lemma which can easily be proved from the definition 2.1 of

the two graph operations.

Lemma 2.1. Let G1, G2, G3 and G4 be four graphs with|V(Gi)|=ni, |E(Gi)|=mi where

i=1,2,3,4. Then,

d_{(}_{G}_{1}

k(G3G2))(u) =

d_{G1}(u) +n_{2} i f u∈V(G_{1})

d_{G3}(u_{3}) +d_{G2}(u_{2}) +n_{1} i f u= (u_{3},u_{2})∈V(G_{3}G_{2})

and

d_{((}_{G4}_{G1}_{)}_{}_{k}_{(}_{G3}_{G2}_{))}(u) =

d_{G}_{4}(u_{4}) +d_{G}_{1}(u_{1}) +n_{2} i f u= (u_{4},u_{1})∈V(G_{4}G_{1})

dG3(u3) +dG2(u2) +n1 i f u= (u3,u2)∈V(G3G2).

For an example we consider G1 =K2, G2 =P2 and G3 =G4 =C4 and hence obtain the

graphsK_{24}(C_{4}P_{3})and(C_{4}×K_{2})_{4}(C_{4}P_{3}), which are shown in figure 1. It is clear from
the definitions of the two operations that |E(G_{1k}(G_{3}G2))|=k(m1+m2+n1n2) +n2m3,
|E((G_{4}G_{1})_{k}(G_{3}G_{2}))| = k(m_{1}+n_{1}n_{2}+m_{2}) +n_{1}m_{4}+m_{3}n_{2} and |V(G_{1k}(G_{3}G_{2}))|=
|V((G_{4}G1)k(G3G2))|=k(n1+n2). Also it is to be exclusively mentioned that|V(G3)|=
|V(G_{4})|=kbutG_{3}6=G_{4}in general.

Here we highlight a standard lemma and two known results without proof. These results will

be expedited in the coming section.

Lemma 2.2. (AM-GM Inequality) Letx_{1},x_{2}, . . . ,x_{n}be nonnegative numbers. Then
x1+x2+. . .+xn

n ≥

n

√

x1x2. . .xn,

where equality holds iff all thex_{i}’s all equal.

Theorem 2.1. [12]Let G_{1},G_{2}, ...,G_{n}be graphs with V_{i}=V(G_{i})and E_{i}=E(G_{i}),1≤i≤n,and
V =V(_{}n_{i}_{=}_{1}G_{i}). Then M_{1}(_{}n_{i}_{=}_{1}G_{i}) =|V|∑n_{i}_{=}_{1}M1_{|}_{V}(Gi)

i| +4|V|∑

n i6=j,j=1

Theorem 2.2. [12]Let G_{1},G_{2}, ...,G_{n}be graphs with V_{i}=V(G_{i})and E_{i}=E(G_{i}),1≤i≤n,and
V =V(_{}n_{i}_{=}_{1}G_{i}) and E =E(_{}n_{i}_{=}_{1}G_{i}). Then M_{2}(_{}n_{i}_{=}_{1}G_{i}) =|V|∑ni=1(

M2(Gi)

|Vi| +3M1(Gi)( |E| |Vi|− |V||Ei|

|Vi|2 )) +4|V|∑

n

i,j,k=1i6=j,i6=k,j6=k

|Ei||Ej||Ek| |Vi||Vj||Vk|.

### 3. Main Results

Theorem 2.3. Let G_{1},G_{2} and G_{3} be graphs with |V_{i}|=|V(G_{i})|=n_{i}, |E_{i}|=|E(G_{i})|=m_{i},
1≤i≤3and n3=k. Then

### ∏

1

(G_{1k}(G_{3}G_{2}))≤

M1(G1) +4n2m1+n2_{2}n1

n_{1}

n1k

×

kM_{1}(G_{2}) +n_{2}M_{1}(G_{3}) +8m_{2}m_{3}+n2_{1}n_{2}k+4n_{1}km_{2}+4n_{1}n_{2}m_{3}
n2k

n2k

,

and

### ∏

2

(G_{1k}(G_{3}G_{2}))≤

n_{2}M_{1}(G_{1}) +M_{2}(G_{1}) +n2_{2}m_{1}
m_{1}

m1k

×

γ1

n_{1}n_{2}k
n1n2k

×

(3m_{3}+n_{1}k)M_{1}(G_{2}) + (3m_{2}+n_{1}n_{2})M_{1}(G_{3}) +kM_{2}(G_{2}) +n_{2}M_{2}(G_{3}) +γ2

β

β

,

whereγ1=8n1m2m3+n21n2m3+n21km2,γ2= (n2m3+km2)(4m1+2n1n2) +k(n21n22+2m1n1n2)

andβ =km2+n2m3. Moreover equalities hold iff all G1, G2and G3are regular graphs.

Proof. Since the set of vertices of G_{1k}(G_{3}G2) can be divided into two categories viz.,

u∈V(G_{1})andu∈V(G_{3}G_{2}), so we have

### ∏

1

(G_{1k}(G_{3}G_{2})) =

_{∏}

u∈V(G1k(G3G2))
d2

G1k(G3G2)(u)

=

_{∏}

u∈V(G1)

d2

G1k(G3G2)(u) !k

×

_{∏}

u=(u3,u2)∈V(G3G2)

d2

G1k(G3G2)(u). (1)

### ∏

1

(G_{1k}(G_{3}G_{2})) =

_{∏}

u∈V(G1)
(d_{G1}(u) +n_{2})2

!k

×

### ∏

u=(u3,u2)∈V(G3G2)

(d_{G3}(u3) +dG2(u1) +n1))2.

AsdG3G2(u3,u2) =dG3(u3) +dG3(u2)the last expression is

=

_{∏}

u∈V(G1)

(d_{G1}(u) +n_{2})2

!k

×

_{∏}

u∈V(G3G2)

(d_{G3}_{G2}(u) +n_{1})2

(2)

and by lemma 2.2 we can write expression ( 2) as

### ∏

1

(G_{1k}(G_{3}G2))≤

∑u∈V(G1)(dG21(u) +2n2dG1(u) +n 2 2)

n_{1}

!n1k

× ∑u∈V(G3G2)(d 2

G3G2(u) +2n1dG3G2(u) +n 2 1

n2k

!n2k

. (3)

Now

(4)

_{∑}

u∈V(G1)

(d_{G1}2 (u) +2n_{2}d_{G1}(u) +n2_{2}) =M_{1}(G_{1}) +4n_{2}m_{1}+n2_{2}n_{1},

and

(5)

_{∑}

u∈V(G3G2)

(d2_{G3}_{G2}(u) +2n1dG3G2(u) +n
2

1) =M1(G3G2) +n21n2k+4n1|E(G3G2)|.

Using theorem 2.1 and the fact that|E(G_{3}G_{2})|=km_{2}+n_{2}m_{3}, we write the expression ( 5)
as

### ∑

u∈V(G3G2)

(d_{G3}2 _{G2}(u) +2n1dG3G2(u) +n
2

1) =n2M1(G3) +kM1(G2) +8m2m3+n21n2k
+4n_{1}km_{2}+4n_{1}n_{2}m_{3}.

(6)

From ( 3), ( 4) and ( 6), we have the first inequality.

Moreover equality holds iff

d_{G1}2 (u_{i}) +2n_{2}d_{G1}(u_{i}) +n2_{2}=d_{G1}2 (u_{j}) +2n_{2}d_{G1}(u_{j}) +n_{2}2 (u_{i},u_{j}∈V(G_{1}))
=⇒ (d_{G1}(u_{i})−d_{G1}(u_{j}))(d_{G1}(u_{i}) +d_{G1}(u_{j}) +2n_{2}) =0

and

d2_{G3}_{G2}(u) +2n_{1}d_{G}_{3}_{G}_{2}(u) +n2_{1}=d_{G3}2 _{G2}(u0) +2n_{1}d_{G}_{3}_{G}_{2}(u0) +n_{1}2 (u,u0∈V(G_{3}G_{2}))
=⇒ (d_{G3}_{G2}(u)−d_{G3}_{G2}(u0))(d_{G3}_{G2}(u) +d_{G3}_{G2}(u0) +2n_{1}) =0.

(8)

So from ( 7) and ( 8), we can conclude that equality holds iff bothG_{1}andG_{3}G_{2}are regular.
Since the set of edges of G_{1k}(G_{3}G_{2}) can again be divided into three categories viz.,
uv∈E(G1),uv∈E(G3G2), anduv∈E(G1k(G3G2))s.t. u∈V(G1)andv∈V(G3G2),

so we have

### ∏

2

(G_{1k}(G_{3}G2)) =

### ∏

uv∈E(G1)(dG1(u) +n2)(dG1(v) +n2)

!k

×

_{∏}

uv∈E(G3G2)

(d_{G3}_{G2}(u) +n_{1})(d_{G3}_{G2}(v) +n_{1})

×

_{∏}

uv∈E(G1k(G3G2),u∈V(G1)and v∈V(G3G2)

(d_{G1}(u) +n_{2})(d_{G3}_{G2}(v) +n_{1})

(9)

Again using lemma 2.1, expression ( 9) can be written as

### ∏

2

(G_{1k}(G_{3}G2))≤

∑uv∈E(G1)(dG1(u) +n2)(dG1(v) +n2)

m_{1}

m1k

×

∑uv∈E(G3G2)(dG3G2(u) +n1)(dG3G2(v) +n1)

km_{2}+n_{2}m_{3}

km2+n2m3

×

∑uv∈E(G1k(G3G2)(dG1(u) +n2)(dG3G2(v) +n1) n1n2k

n1n2k

(10)

Now

(11)

_{∑}

uv∈E(G1)

(dG1(u) +n2)(dG1(v) +n2) =k M2(G1) +n2M1(G1) +n 2 2m1

,

also using theorem 2.1 and theorem 2.2, we have

∑uv∈E(G3G2)(dG3G2(u) +n1)(dG3G2(v) +n1)

=kM_{2}(G_{2}) +n_{2}M_{2}(G_{3}) + (3m_{3}+n_{1}k)M_{1}(G_{2}) + (3m_{2}+n_{1}n_{2})M_{1}(G_{3})
+8n_{1}m_{2}m_{3}+n_{1}2n_{2}m_{3}+n2_{1}km_{2},

and

### ∑

uv∈E(G1k(G3G2)

(dG1(u) +n2)(dG3G2(v) +n1) =(n2m3+km2)(4m1+2n1n2)

+k(n2_{1}n2_{2}+2m1n1n2).

(13)

Using ( 11), ( 12), ( 13) in ( 10), we have the second inequality.

From lemma 2.2 for connected graphsG1,G2andG3equality holds iff (dG1(u) +n2)(dG1(v) +n2) = (dG1(u) +n2)(dG1(v

0_{) +}_{n}

2) (for anyuv,uv0∈E(G1)),

for anyuv,uv0∈E(G_{3}G_{2}),

d_{G3}_{G2}(u) +n_{1})(d_{G3}_{G2}(v) +n_{1}=d_{G3}_{G2}(u) +n_{1})(d_{G3}_{G2}(v0) +n_{1},
and for anyuv,uv0∈E(G_{1k}(G_{3}G2),u∈V(G1)andv,v0∈V(G3G2),

dG1(u) +n2)(dG3G2(v) +n1=dG1(u) +n2)(dG3G2(v

0_{) +}_{n}

1.

Hence equality holds in the second inequality iff all the graphsG_{1},G_{2}andG_{3}are connected

regular graphs.

Theorem 2.4. Let G_{1},G_{2}, G_{3}and G_{4} be graphs with|V_{i}|=|V(G_{i})|=n_{i},|E_{i}|=|E(G_{i})|=m_{i},
1≤i≤3and n3=k=n4. Then

### ∏

1

((G_{4}G_{1})_{k}(G_{3}G_{2}))≤

kM_{1}(G_{1}) +n_{1}M_{1}(G_{4}) +8m_{1}m_{4}+4n_{2}(km_{1}+n_{1}m_{4}) +n2_{2}kn_{1}
n1k

n1k

×

_{kM}

1(G2) +n2M1(G3) +8m2m3+4n1(km2+n2m3) +n21kn2

n_{2}k

n2k

,

and

∏2((G4G1)k(G3G2)) ≤

kM_{2}(G_{1}) +n_{1}M_{2}(G_{4}) +3m_{4}M_{1}(G_{1}) +3m_{1}M_{1}(G_{4}) +n_{2}(kM_{1}(G_{1}) +n_{1}M_{1}(G_{4}) +δ14)

m_{1}k+m_{4}n_{1}

m1k+m4n1

kM_{2}(G_{2}) +n_{2}M_{2}(G_{3}) +3m_{3}M_{1}(G_{2}) +3m_{2}M_{1}(G_{3}) +n_{1}(kM_{1}(G_{2}) +n_{2}M_{1}(G_{3}) +δ23)

m_{2}k+m_{3}n_{2}

m2k+m3n2

× γ2+n1n2∑ k

i=1dG4(u4i)dG3(vi3)

n1n2k

!n1n2k

where ui_{4}∈V(G_{4}), vi_{3}∈V(G_{3})and for fixed i,(u_{4}i,u_{1})is adjacent to(vi_{3},v_{2})for any u_{1}∈V(G_{1}),
v2 ∈V(G2) and δpq = 8mpmq+npn2mq+knqmp, γ2 = 4n1m2m4+2n_{1}2n2m4+4n2m1m3+

4m_{1}m_{2}k+2kn_{1}m_{1}n_{2}+2n_{1}n2_{2}m_{3}+2kn_{1}n_{2}m_{2}+kn2_{1}n2_{2}. Moreover equalities hold iff all G_{1}, G_{2},
G3and G4are regular graphs.

Proof. There are three two types of vertices present in (G_{4}G1)k(G3G2), they are u∈

G_{4}G_{1}andu∈G_{3}G_{2}.

### ∏

1

((G_{4}G1)k(G3G2)) =

### ∏

u∈V((G4G1)k(G3G2))

d_{(}2_{G4}_{G1}_{)}_{}

k(G3G2)(u)

=

_{∏}

u=(u3,u2)∈V(G3G2)

d_{(}2

G4G1)k(G3G2)(u)

×

_{∏}

u=(u4,u1)∈V(G3G2)

d_{(}2_{G4G1}_{)}_{}

k(G3G2)(u), (14)

using lemma 2.1 and 2.2, we can write ( 14) as

### ∏

1

((G_{4}G1)k(G3G2))≤

∑u∈V(G4G1)(d(G4G1)(u) +n2) 2

n_{1}k

!n1k

× ∑u∈V(G3G2)(d(G3G2)(u) +n1) 2

n2k

!n2k

. (15)

Now

### ∑

u∈V(G4G1)

(d_{(}_{G4}_{G1}_{)}(u) +n_{2})2=kM_{1}(G_{1}) +n_{1}M_{1}(G_{4}) +8m_{1}m_{4}+4n_{2}(km_{1}+n_{1}m_{4})

+n2_{2}kn_{1},
and

### ∑

u∈V(G3G2)

(d_{(}_{G3}_{G2}_{)}(u) +n_{1})2=kM_{1}(G_{2}) +n_{2}M_{1}(G_{3}) +8m_{2}m_{3}+4n_{1}(km_{2}+n_{2}m_{3})

+n2_{1}kn_{2}.
Hence we have the first inequality of the theorem.

For equality we must have

(16) (d_{(}_{G}_{4}_{G}_{1}_{)}(u) +n2)2= (d(G4G1)(v) +n2)

2 _{(}_{for any}_{uv}_{∈}_{E}_{(}_{G}

and

(17) (d_{(}_{G3}_{G2}_{)}(u) +n_{1})2= (d_{(}_{G3}_{G2}_{)}(v) +n_{1})2 (for anyuv∈E(G_{3}G_{2})).

From ( 16) and ( 17) it is easy to show that the equality holds iff all the graphsG1, G2, G3

andG_{4}are regular.

Now the set of edges of(G_{4}G1)k(G3G2)can again classified into three categories viz.,

uv∈E(G_{4}G_{1}), uv∈E(G_{3}G_{2}), anduv∈E((G_{4}G_{1})_{k}(G_{3}G_{2}))whereu∈V(G_{4}G_{1}),
v∈V(G_{3}G2). Hence we have

### ∏

1

((G_{4}G_{1})_{k}(G_{3}G_{2})) =

_{∏}

uv∈E(G4G1)
(d_{(}_{G}_{4}_{G}

1)(u) +n2)(d(G4G1)(v) +n2)

×

_{∏}

uv∈E(G3G2)

(d_{(}_{G3}_{G2}_{)}(u) +n_{1})(d_{(}_{G3}_{G2}_{)}(v) +n_{1})

×

_{∏}

uv∈E((G4G1)k(G3G2)

(d_{(}_{G4}_{G1}_{)}(u) +n_{2})(d_{(}_{G3}_{G2}_{)}(v) +n_{1})

(18)

Again using lemma 2.2, we have

### ∏

1

((G_{4}G_{1})_{k}(G_{3}G_{2}))≤

∑uv∈E(_{G4}G1)(d(_{G4}G1)(u) +n2)(d(_{G4}G1)(v) +n2)

n1m4+km1

n1m4+km1

×

∑uv∈E(G3G2)(d(G3G2)(u) +n1)(d(G3G2)(v) +n1)

n_{2}m_{3}+m_{2}k

n2m3+m2k

×

∑uv∈E((G4G1)k(G3G2)(d(G4G1)(u) +n2)(d(G3G2)(v) +n1)

kn_{1}n_{2}

kn1n2

. (19)

Hence we have the second inequality of the theorem from the following expressions.

### ∑

uv∈E(G4G1)

(d_{(}_{G4}_{G1}_{)}(u) +n_{2})(d_{(}_{G4}_{G1}_{)}(v) +n_{2}) =kM_{2}(G_{1}) +n_{1}M_{2}(G_{4}) +3m_{4}M_{1}(G_{1})

### ∑

uv∈E(G3G2)

(d_{(}_{G3}_{G2}_{)}(u) +n_{1})(d_{(}_{G3}_{G2}_{)}(v) +n_{1}) =kM_{2}(G_{2}) +n_{2}M_{2}(G_{3}) +3m_{3}M_{1}(G_{2})

+3m_{2}M_{1}(G_{3}) +n_{2}(kM_{1}(G_{2}) +n_{1}M_{1}(G_{3})
+8m_{2}m_{3}+n_{1}n_{2}m_{3}+km_{2}),

and finally letu∈V(G_{4}G_{1})andv∈V(G_{3}G_{2})s.t.u= (ui_{4},u_{1}),v= (v_{3}i,v_{2}), whereV(G_{3}) =
{u1_{3},u2_{3}, ...,u_{3}k}andV(G_{4}) ={v1_{4},v2_{4}, ...,vk_{4}}. Also for fixedi,uis adjacent tovfor allu_{1}∈V(G_{1})

andv_{2}∈V(G_{2}). Then we have

### ∑

uv∈E((G4G1)k(G3G2))

d(u)d(v) = k

### ∑

i=1u2∈

### ∑

V(G2)v1∈### ∑

V(G1)(d_{G3}(ui_{3}) +d_{G2}(u_{2}) +n_{1})(d_{G4}(ui_{4})+

d_{G1}(v1) +n2)

=4n1m2m4+2n21n2m4+4n2m1m3+4m1m2k+2kn1m1n2

+2n_{1}n2_{2}m_{3}+2kn_{1}n_{2}m_{2}+kn2_{1}n2_{2}+n_{1}n_{2}

k

### ∑

i=1

d_{G}_{4}(ui_{4})d_{G}_{3}(vi_{3}).

For equality of the second result in the theorem, we must have

for anyuv uv0∈E(G_{4}G_{1})

(20) (d_{(}_{G4}_{G1}_{)}(u) +n_{2})(d_{(}_{G4}_{G1}_{)}(v) +n_{2}) = (d_{(}_{G4}_{G1}_{)}(u) +n_{2})(d_{(}_{G4}_{G1}_{)}(v0) +n_{2}),

and for anyuv uv0∈E(G_{3}G_{2})

(21) (d_{(}_{G}_{3}_{G}

2)(u) +n1)(d(G3G2)(v) +n1) = (d(G3G2)(u) +n1)(d(G3G2)(v

0_{) +}_{n}

1)

From expression ( 20) and ( 21), we can easily see that equality holds iff all the graphsG1,

G_{2},G_{3}andG_{4}are regular. _{}

Conflict of Interests

The authors declare that there is no conflict of interests.

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