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)

#pragma GCC optimize("O3")
#define _CRT_SECURE_NO_WmarkNINGS
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define M 1000000007
#define M2 998244353
#define ll long long
#define pll pair<long, long>
#define REP(i, a, b) for (ll i = a; i < b; i++)
#define REPI(i, a, b) for (ll i = b - 1; i >= a; i--)
#define ff first
#define ss second
#define pb push_back
#define db pop_back
#define mp make_pair
#define mt make_tuple
#define g(a, b) get<a>(b)
#define INF (ll)1e18 + 100
#define vl vector<ll>
#define vi vector<int>
#define vll vector<pair<ll, ll>>
#define vii vector<pair<int, int>>
#define all(v) v.begin(), v.end()
#define bset(a, p) ((a) | (1ll << (p)))
#define bchk(a, p) ((a) & (1ll << (p)))
#define bxor(a, p) ((a) ^ (1ll << (p)));
#define brem(a, p) (bchk(a, p) ? (bxor(a, p)) : (a))
/*SOME BITMASK KNOWLEDGE
1)x & (x - 1):sets the last one bit of x to zero
power of two exactly when x & (x − 1) = 0.
2)x & -x:sets all the one bits to zero, except last one bit
3)x | (x - 1):inverts all the bits after the last one bit*/
#define endl "\n"
#define p0(a) cout << a << " "
#define p1(a) cout << a << endl
#define p2(a, b) cout << a << " " << b << endl
#define p3(a, b, c) cout << a << " " << b << " " << c << endl
#define p4(a, b, c, d) cout << a << " " << b << " " << c << " " << d << endl

#define o_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define o_setll tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
//member functions :
//1. order_of_key(k) : number of elements strictly lesser than k
//2. find_by_order(k) : k-th element in the set

ll modI(ll a, ll m);
ll gcd(ll a, ll b);
ll powM(ll x, unsigned ll y, unsigned ll m);
void pairsort(int a[], int b[], int n);
void pairsortll(ll a[], ll b[], ll n);
ll logint(ll x, ll y);

ll gcd(ll x, ll y)
{
    if (x == 0)
        return y;
    return gcd(y % x, x);
}

ll powM(ll x, ll y, ll m)
{
    if (y == 0)
        return 1;
    ll p = powM(x, y / 2, m) % m;
    p = (p * p) % m;
    return (y % 2 == 0) ? p : (x * p) % m;
}

/*ll modI(ll a, ll m)
{
   return powM(a, m-2, m);
}*/
ll modI(ll a, ll m)
{
    ll m0 = m, y = 0, x = 1;
    if (m == 1)
        return 0;
    while (a > 1)
    {
        ll q = a / m;
        ll t = m;
        m = a % m;
        a = t;
        t = y;
        y = x - q * y;
        x = t;
    }
    if (x < 0)
        x += m0;
    return x;
}
void pairsort(int a[], int b[], int n)
{
    pair<int, int> v[n];
    REP(i, 0, n)
    {
        v[i].ff = a[i];
        v[i].ss = b[i];
    }
    sort(v, v + n);
    REP(i, 0, n)
    {
        a[i] = v[i].ff;
        b[i] = v[i].ss;
    }
}

void pairsortll(ll a[], ll b[], ll n)
{
    pair<ll, ll> v[n];
    REP(i, 0, n)
    {
        v[i].ff = a[i];
        v[i].ss = b[i];
    }
    sort(v, v + n);
    REP(i, 0, n)
    {
        a[i] = v[i].ff;
        b[i] = v[i].ss;
    }
}

ll logint(ll x, ll y)
{
    ll ans = 0;
    ll a = 1;
    for (ll i = 0; i <= x; i++)
    {
        if (x < a)
        {
            return ans;
        }
        ans++;
        a *= y;
    }
    return -1;
}
const int N = 2e5+5;
ll dist1[N], dist2[N];
vector<ll>v[N];
bool vis1[N], vis2[N];

double f(double b, double c, double x){
    return (x*x+b*x+c)/sin(x);
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    //clock_t begin = clock();
    int ar[3][3];
    pair<int,int>p[105];
    int mark[3][3];
    memset(mark,0,sizeof(mark));
    map<int,int>m;
    REP(i,0,3){
        REP(j,0,3){
            cin>>ar[i][j];
            m[ar[i][j]]++;
            p[ar[i][j]] = {i,j};
        }
    }
    
    int t;
    cin>>t;
    REP(i,0,t){
        int x;
        cin>>x;
        if(m[x] == 0){continue;}
        mark[p[x].ff][p[x].ss]=1;
    }
    int flag = 0;
    REP(i,0,3){
        if(mark[i][0] && mark[i][1] && mark[i][2]){flag=1;}
        if(mark[0][i] && mark[1][i] && mark[2][i]){flag=1;}
    }
    if(mark[0][0] && mark[1][1] && mark[2][2]){flag=1;}
    if(mark[0][2] && mark[1][1] && mark[2][0]){flag=1;}
    if(flag){
        cout<<"Yes";
    }
    else{
        cout<<"No";
    }

}