Problem: You are given a correct solution of the sudoku puzzle. If you don't know what is the sudoku, you can read about it [here](http://tiny.cc/636xmz).

The picture showing the correct sudoku solution:

<image>

Blocks are bordered with bold black color.

Your task is to change at most 9 elements of this field (i.e. choose some 1 ≤ i, j ≤ 9 and change the number at the position (i, j) to any other number in range [1; 9]) to make it anti-sudoku. The anti-sudoku is the 9 × 9 field, in which:

  * Any number in this field is in range [1; 9]; 
  * each row contains at least two equal elements; 
  * each column contains at least two equal elements; 
  * each 3 × 3 block (you can read what is the block in the link above) contains at least two equal elements. 



It is guaranteed that the answer exists.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

Each test case consists of 9 lines, each line consists of 9 characters from 1 to 9 without any whitespaces — the correct solution of the sudoku puzzle.

Output

For each test case, print the answer — the initial field with at most 9 changed elements so that the obtained field is anti-sudoku. If there are several solutions, you can print any. It is guaranteed that the answer exists.

Example

Input


1
154873296
386592714
729641835
863725149
975314628
412968357
631457982
598236471
247189563


Output


154873396
336592714
729645835
863725145
979314628
412958357
631457992
998236471
247789563

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
char fek(long long n) {
  if (n == 9) {
    return '1';
  } else {
    return char('0' + n + 1);
  }
}
signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  long long t;
  cin >> t;
  for (long long _ = 0; _ < (long long)t; ++_) {
    vector<string> a(9);
    for (long long i = 0; i < (long long)9; ++i) {
      cin >> a[i];
    }
    a[0][0] = fek(a[0][0] - '0');
    a[1][3] = fek(a[1][3] - '0');
    a[2][6] = fek(a[2][6] - '0');
    a[3][1] = fek(a[3][1] - '0');
    a[4][4] = fek(a[4][4] - '0');
    a[5][7] = fek(a[5][7] - '0');
    a[6][2] = fek(a[6][2] - '0');
    a[7][5] = fek(a[7][5] - '0');
    a[8][8] = fek(a[8][8] - '0');
    for (long long i = 0; i < (long long)9; ++i) {
      cout << a[i] << endl;
    }
    cout << endl;
  }
  return 0;
}