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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

#define L 3

int main(void) {
  int A[L][L];
  int i,j,k;
  int N;
  int b[10];
  int sum;
  int bingo = 0;

  // ビンゴカード
  for (i=0; i<L; i++) {
    for (j=0; j<L; j++) {
      scanf("%d", &A[i][j]);
    }
  }

  // 数字選び
  scanf("%d", &N);
  for (k=0; k<N; k++) {
    scanf("%d", &b[k]);
  }

  // 印付け
  for (i=0; i<L; i++) {
    for (j=0; j<L; j++) {
      for (k=0; k<N; k++) {
        if (A[i][j] == b[k]) {
          A[i][j] = 0;
          break;
        }
      }
    }
  }

  // ビンゴ確認
  // for (i=0; i<L; i++) {
  //   for (j=0; j<L; j++) {
  //     printf("%d ", A[i][j]);
  //   }
  //   printf("\n");
  // }
  // printf("\n");

  for (i=0; i<L; i++) {
    sum = 0;
    for (j=0; j<L; j++) {
      sum += A[i][j];
    }
    // printf("sum: %d\n", sum);
    if (sum == 0) bingo = 1;
  }
  for (j=0; j<L; j++) {
    for (i=0; i<L; i++) {
      sum = 0;
      sum += A[i][j];
    }
    if (sum == 0) bingo = 1;
  }
  if (A[0][0] + A[1][1] + A[2][2] == 0) bingo = 1;
  if (A[0][2] + A[1][1] + A[2][0] == 0) bingo = 1;

  if (bingo) printf("Yes");
  else printf("No");
  printf("\n");
  return (0);
}