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 <cstdio>
#include <iostream>
#include <string>
#include <vector>
#include <array>
#include <algorithm>
#include <utility>
#include <numeric>
#include <deque>
#include <stack>
#include <queue>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <cmath>
#include <limits>
#include <cassert>
#include <bitset>
#include <list>
#include <iomanip>
#include <cctype>

#define for0(i, N)    for (ll i = 0  ; i <  (N); ++i)
#define for1n(i, N)   for (ll i = 1  ; i <= (N); ++i)
#define for1(i, N)    for (ll i = 1  ; i <  (N); ++i)
#define forn(i, N)    for (ll i = 0  ; i <= (N); ++i)
#define forx(i, N, x) for (ll i = (x); i <  (N); ++i)
#define forl(i, N, x, a) for (ll i = (x), a = 0; a < (N);   \
                              i = (i + 1) % (N), ++a)
#define int long long
typedef long long ll;


template<size_t i>
std::bitset<i> operator++(std::bitset<i> b, int){
    static_assert(i <= 64, "Cannot cast 64bit int.");
    return b = (unsigned long long)b + 1;
}

template<size_t i>
std::bitset<i> operator++(std::bitset<i> b){
    static_assert(i <= 64, "Cannot cast 64bit int.");
    auto temp = b;
    b++;
    return temp;
}

template<size_t i>
std::bitset<i> operator--(std::bitset<i> b, int){
    static_assert(i <= 64, "Cannot cast 64bit int.");
    return b = (unsigned long long)b - 1;
}

template<size_t i>
std::bitset<i> operator--(std::bitset<i> b){
    static_assert(i <= 64, "Cannot cast 64bit int.");
    auto temp = b;
    b--;
    return temp;
}

ll constexpr inf = std::numeric_limits<ll>::max();

using namespace std;

typedef long double ld;
typedef priority_queue<ll> pq;
typedef priority_queue<ll,vector<ll>, greater<ll>> pql;
typedef stack<ll> stk;
typedef queue<ll> qu;
typedef pair<ll, ll> pll;
typedef map<ll, ll> mll;
typedef unordered_map<ll, ll> umll;
typedef set<ll> sll;
typedef unordered_set<ll> usll;
typedef vector<ll> vll;
typedef vector<ld> vld;
typedef vector<vector<ll>> vvll;


signed main(void){
    cin.tie(0); ios::sync_with_stdio(false);cout<<fixed<<setprecision(10);
    vector<vll> a(3, vll(3, 0));

    for0(i, 3){
        for0(j, 3)
            cin >> a[i][j];
    }
    int n;
    cin >> n;

    vector<vector<bool >> q(3, vector<bool>(3, false));

    for0(t, n){
        int b;
        cin >> b;

        for0(i, 3)for0(j, 3){
            if(a[i][j] == b){
                q[i][j] = true;
                goto out;
            }
        out:
            continue;
        }
    }

    // tate
    for0(i, 3){
        bool good = false;
        for0(j, 3){
            good = good && q[i][j];
        }
        if (good) {
            cout << "Yes";
            return 0;
        }
    }

    // yoko
    for0(i, 3){
        bool good = false;
        for0(j, 3){
            good = good && q[j][i];
        }
        if (good) {
            cout << "Yes";
            return 0;
        }
    }

    // naname
    if ((q[0][0] && q[1][1] && q[2][2]) || (q[0][2] && q[1][1] && q[2][0])) {
        cout << "Yes";
        return 0;
    }

    cout << "No";

    return 0;
}