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 <bits/stdc++.h>
#include <type_traits>
using namespace std;

#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P)
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, vector<vector<T> > P)
{ for (int i = 0; i < P.size(); ++i) { s << endl << P[i]; } return s << endl; }
template<class T> ostream& operator << (ostream &s, set<T> P)
{ for(auto it : P) { s << "<" << it << "> "; } return s << endl; }
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P)
{ for(auto it : P) { s << "<" << it.first << "->" << it.second << "> "; } return s << endl; }

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rrep(i, n) for (int i = 1; i <= (n); ++i)
#define drep(i, n) for (int i = (n)-1; i >= 0; --i)
#define ddrep(i, n) for (int i = n; i > 0; --i)
#define srep(i, s, t) for (int i = s; i < t; ++i)
#define ssrep(i, s, t) for (int i = s; i <= t; ++i)

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

using ll = long long;
using pi = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using ld = long double;
using str = string;

#define rng(a) a.begin(), a.end()
#define rrng(a) a.rbegin(), a.rend()
#define pb push_back
#define eb emplace_back
#define fi first
#define se second

long int M = 1e+9+7;
int inf = 1e+9;

int main(void){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << std::setprecision(10);

    int ans=0;
    vvi a(3,vi(3));
    vvi flag(3,vi(3));
    rep(i,3){
        rep(j,3){
            cin >> a[i][j];
        }
    }
    int n;
    cin >> n;
    vi b(n);
    rep(i,n){
        cin >> b[i];
        rep(j,3){
            rep(k,3){
                if(b[i] == a[j][k]) flag[j][k] = 1;
            }
        }
    }

    rep(i,3){
        if(flag[i][0] && flag[i][1] && flag[i][2]) ans++;
        if(flag[0][i] && flag[1][i] && flag[2][i]) ans++;
    }
    if(flag[0][0] && flag[1][1] && flag[2][2]) ans++;
    if(flag[0][2] && flag[1][1] && flag[2][0]) ans++;

    /*
    rep(i,3){
        rep(j,3){
            cout << flag[i][j] << " ";
        }
        cout << endl;
    }
    */

    if(ans) cout << "Yes" << endl;
    else cout << "No" << endl;
    return 0;
}