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() {
  int t;
  scanf("%d", &t);
  for (int tt = 0; tt < t; tt++) {
    vector<vector<int> > A(9, vector<int>(9));
    for (int i = 0; i < 9; i++) {
      int q;
      scanf("%d", &q);
      for (int j = 0; j < 9; j++) {
        A[i][8 - j] = q % 10;
        q /= 10;
      }
    }
    A[0][0] = A[0][1];
    A[1][3] = A[0][3];
    A[2][6] = A[0][6];
    A[3][1] = A[3][0];
    A[4][4] = A[3][4];
    A[5][7] = A[4][7];
    A[6][2] = A[6][1];
    A[7][5] = A[8][5];
    A[8][8] = A[7][7];
    for (int i = 0; i < 9; i++) {
      for (int j = 0; j < 9; j++) printf("%d", A[i][j]);
      printf("\n");
    }
  }
  return 0;
}