

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
#include <bits/stdc++.h>
#pragma GCC optimize ("Ofast")
#define ll long long
#define pll pair<long long,long long>
#define pii pair<int,int>
#define pb push_back
#define F first
#define S second
#define forn(i, n) for(int i=0; i<int(n); i++)
#define Forn(i, n) for(int i=1; i<=int(n); i++)

using namespace std;

int a[30][5] = {1,0,0,1,0,
1,1,0,2,0,
2,0,0,1,1,
2,1,0,1,2,
1,1,0,1,1,
2,1,0,2,1,
2,2,0,2,2,
1,2,0,2,1,
1,1,0,1,1,
1,2,0,1,2,
1,0,1,2,0,
1,1,1,3,0,
2,0,1,2,1,
2,1,1,2,2,
1,1,1,2,1,
2,1,1,3,1,
2,2,1,3,2,
1,2,1,3,1,
1,1,1,2,1,
1,2,1,2,2,
1,0,2,2,1,
1,1,2,3,1,
1,2,1,1,3,
2,0,2,2,2,
2,1,2,2,3,
1,1,2,2,2};

int b[5];

char fnd()
{
    forn(i, 26)
    {
        int sum = 0;
        forn(j, 5) if(a[i][j] == b[j]) sum++;
        if(sum == 5) return (char)('a'+i);
    }
    return '!';
}

int main()
{
    cin.tie(0); ios_base::sync_with_stdio(false);
    int n;
    cin >> n;
    while(n--)
    {
        forn(i, 5) cin >> b[i];
        cout << fnd();
    }
    cout << endl;
}
