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`.

Examples

Input

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


Output

Yes


Input

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


Output

No


Input

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


Output

Yes
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++) 
using ll = long long;


int main() {

	int a[3][3], n, b[10];
	bool ab[3][3], ans = false;
	rep(i, 3) rep(j, 3) ab[i][j] = false;

	rep(i, 3) rep(j, 3) cin >> a[i][j];
	cin >> n;
	rep(i, n) cin >> b[i];

	rep(i, 3) rep(j, 3)rep(k, n) if (b[k] == a[i][j])ab[i][j] = true;




	rep(i, 3)if (ab[i][0] & ab[i][1] & ab[i][2]) ans = true;
	rep(j, 3)if (ab[0][j] & ab[1][j] & ab[2][j]) ans = true;

	if (ab[0][0] & ab[1][1] & ab[2][2]) ans = true;
	if (ab[0][2] & ab[1][1] & ab[2][0]) ans = true;


	if (ans == true)cout << "Yes" << endl;
	else cout << "No" << endl;
}