Problem:
Score : 200 points

Problem Statement

We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}.

The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet.

Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal.

Constraints

All values in input are integers.

1 \leq A_{i, j} \leq 100

A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2))

1 \leq N \leq 10

1 \leq b_i \leq 100

b_i \neq b_j (i \neq j)

Input

Input is given from Standard Input in the following format:

A_{1, 1} A_{1, 2} A_{1, 3}
A_{2, 1} A_{2, 2} A_{2, 3}
A_{3, 1} A_{3, 2} A_{3, 3}
N
b_1
\vdots
b_N

Output

If we will have a bingo, print Yes; otherwise, print No.

Sample Input 1

84 97 66
79 89 11
61 59 7
7
89
7
87
79
24
84
30

Sample Output 1

Yes

We will mark A_{1, 1}, A_{2, 1}, A_{2, 2}, A_{3, 3}, and complete the diagonal from the top-left to the bottom-right.

Sample Input 2

41 7 46
26 89 2
78 92 8
5
6
45
16
57
17

Sample Output 2

No

We will mark nothing.

Sample Input 3

60 88 34
92 41 43
65 73 48
10
60
43
88
11
48
73
65
41
92
34

Sample Output 3

Yes

We will mark all the squares.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define pll pair<ll,ll>
#define x first
#define y second
#define mod 1e9+7
#define f(i,a,b) for(ll i=a;i<b;i++)
#define rof(i,a,b) for(ll i=a;i>b;i--)

//int a[1001][1001] //aise global me hi

ll lcm(ll a, ll b){
    return a / __gcd(a, b) * b;
}
//memset can initialize to 0 or -1 ONLY //memset(str, 't', sizeof(str)); // string
//memset(a, 0, sizeof(a));  //1D array    // memset(a,0,sizeof(a[0][0])*n*n); //2D array
// set.insert(),find()    // sort(v.begin(),v.end(),compare)

void solve()
{
  int a[3][3];
  f(i,0,3)
  f(j,0,3) cin>>a[i][j];
  bool out=false;
int n;cin>>n;
unordered_map<int,int> m;
while(n--)
{
int x;cin>>x;m[x]++;
}

f(i,0,3)
{bool u=true;
	f(j,0,3)
	{
		if(m.count(a[i][j])==0) u=false; 
	}
	out=u;
}


f(i,0,3)
{bool u=true;
	f(j,0,3)
	{
		if(m.count(a[j][i])==0) u=false; 
	}
	if(u) out=u;
}

if(m.count(a[0][0])!=0&&m.count(a[2][2])!=0&&m.count(a[1][1])!=0)
out=true;
if(m.count(a[1][1])!=0&&m.count(a[0][2])!=0&&m.count(a[2][0])!=0)
out=true;

if(out) cout<<"Yes";
else cout<<"No";


}


int main()
{

ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    ll t=1;
    //cin>>t;
    while(t--)
    {
    
    solve();
    
    }

}