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 n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    string a1, a2, a3, a4, a5, a6, a7, a8, a9;
    cin >> a1 >> a2 >> a3 >> a4 >> a5 >> a6 >> a7 >> a8 >> a9;
    a1[0] = a2[0];
    a4[1] = a5[1];
    a7[2] = a8[2];
    a2[3] = a3[3];
    a5[4] = a6[4];
    a8[5] = a9[5];
    a3[6] = a2[6];
    a6[7] = a5[7];
    a9[8] = a8[8];
    cout << a1 << endl
         << a2 << endl
         << a3 << endl
         << a4 << endl
         << a5 << endl
         << a6 << endl
         << a7 << endl
         << a8 << endl
         << a9 << endl;
  }
  return 0;
}