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;
int main() {
  long long int t, ti = 0;
  cin >> t;
  while ((ti++) != t) {
    long long int ar[10][10];
    for (int i = 1; i <= 9; i++) {
      for (int j = 1; j <= 9; j++) {
        scanf("%1lld", &ar[i][j]);
      }
    }
    ar[2][2] = ar[2][1];
    ar[1][7] = ar[1][6];
    ar[3][6] = ar[3][9];
    ar[5][3] = ar[5][1];
    ar[4][9] = ar[4][6];
    ar[6][5] = ar[6][8];
    ar[8][1] = ar[8][2];
    ar[9][4] = ar[9][3];
    ar[7][8] = ar[7][7];
    for (int i = 1; i <= 9; i++) {
      for (int j = 1; j <= 9; j++) printf("%lld", ar[i][j]);
      printf("\n");
    }
  }
}