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 int m1 = 10;
int s[m1][m1];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    for (int i = 0; i < 9; i++)
      for (int j = 0; j < 9; j++) scanf("%1d", &s[i][j]);
    s[0][0] = (s[0][0] + 1) % 10;
    s[1][3] = (s[1][3] + 1) % 10;
    s[2][6] = (s[2][6] + 1) % 10;
    s[3][1] = (s[3][1] + 1) % 10;
    s[4][4] = (s[4][4] + 1) % 10;
    s[5][7] = (s[5][7] + 1) % 10;
    s[6][2] = (s[6][2] + 1) % 10;
    s[7][5] = (s[7][5] + 1) % 10;
    s[8][8] = (s[8][8] + 1) % 10;
    for (int i = 0; i < 9; i++) {
      for (int j = 0; j < 9; j++) printf("%1d", s[i][j] ? s[i][j] : 1);
      cout << endl;
    }
  }
  return 0;
}