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: Accepted)

#include <iostream>
#include <vector> 
#include <string>
#include <set>
#include <unordered_set>
using namespace std;
 
int main() {
	
     vector<vector<int>>  grid(3, vector<int> (3));
     
     for(int i = 0; i < 3; i++) {
     	for(int j = 0; j < 3; j++) {
     		cin >> grid[i][j];
     	}
     }
     int n;
     cin >> n;
     vector<int> bingo(n);
     for(int i = 0; i < n; i++) cin >> bingo[i];
     
	 unordered_set<int> uset;
	 int count = 0;
	 int row = 3;
	 int col = 3;
	 bool done = false;
	 for(int i = 0; i < n; i++) uset.insert(bingo[i]);
	 for(int i = 0; i < 3; i++) {
     	for(int j = 0; j < 3; j++) {
     		
     		// check column by column
     		for(int k = 0; k < 3; k++ ) {
     			if(uset.find(grid[j][k]) != uset.end()) {
     				count++;
     			}
     		}
     		if(count == 3) {
     		   cout << "Yes" << "\n";
     		   return 0;
     		}
     		count = 0;
     		// to check row by row
     		for(int k = 0; k < 3; k++) {
     			if(uset.find(grid[k][j]) != uset.end()) {
     				count++;
     			}
     		}
     		
     		if(count == 3) {
     		   cout << "Yes" << "\n";
     		   return 0;
     		}
     		// to check top uppper left corder 
     		count= 0;
     		for(int k = 0; k < 3; k++) {
     			if(uset.find(grid[row - k - 1][col- k - 1]) != uset.end()) {
     				count++;
     			}
     		}
     		
     		if(count == 3) {
     		   cout << "Yes" << "\n";
     		   return 0;
     		}
     		
     		count = 0;
     		// to check top upper right corner 
     		for(int k = 0; k < 3; k++) {
     			if(uset.find(grid[k][col - k - 1]) != uset.end()) {
     				count++;
     			}
     		}
     		if(count == 3) {
     		   cout << "Yes" << "\n";
     		   
     		   return 0;
     		}
     		count = 0;
     	}
     }
	 cout << "No" << "\n";
	 
  return 0;
}