Problem: 

Input

The first line of the input contains a single integer N (1 ≤ N ≤ 24). The next N lines contain 5 space-separated integers each. The first three integers will be between 0 and 2, inclusive. The last two integers will be between 0 and 3, inclusive. The sum of the first three integers will be equal to the sum of the last two integers.

Output

Output the result – a string of lowercase English letters.

Examples

Input


1
1 0 0 1 0


Output


a


Input


10
2 0 0 1 1
1 1 1 2 1
2 1 0 1 2
1 1 0 1 1
2 1 0 2 1
1 1 1 2 1
1 2 1 3 1
2 0 0 1 1
1 1 0 1 1
1 1 2 2 2


Output


codeforcez

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL,LL> PLL;
const int INF = 0x3f3f3f3f, N = 1e5 + 10;
inline int lc(int u) {return u << 1;}
inline int rc(int u) {return u << 1 | 1;}
inline int lowbit(int x) {return x & (-x);}
int y[27] = {10010,11020,20011,21012,11011,21021,22022,12021,11011,12012,10120,11130,20121,21122,11121,21131,22132,12131,11121,12122,10221,11231,12113,20222,21223,11222};
char x[30000];
inline void solve() {
    for (int i = 0; i < 26; i ++ ) {
        if (x[y[i]] == 0) x[y[i]] = i + 'a';
    }
    int n; cin >> n;
    for (int i = 1; i <= n; i ++ ) {
        int res = 0;
        for (int j = 0; j < 5; j ++ ) {
            int tmp; cin >> tmp;
            res = res * 10 + tmp;
        }
        cout << x[res];
    }
}
int main() {
//    ios::sync_with_stdio(false), cin.tie(nullptr);
//    int t; cin >> t;
//    while (t -- )
        solve();
    return 0;
}