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 Mod = 1e9 + 7;
bool isPowerOfTwo(int x) { return x && (!(x & (x - 1))); }
long long int pow(long long int x, long long int y, long long int mod) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = (ans * x) % mod;
    y /= 2;
    x = (x * x) % mod;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int test, n, i, j;
  long long int arr[10][10];
  string s;
  n = 9;
  cin >> test;
  while (test--) {
    for (i = 1; i <= n; i++) {
      cin >> s;
      for (j = 0; j < s.length(); j++) {
        arr[i][j + 1] = s[j] - '0';
      }
    }
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        if (i == 1 && j == 7) {
          if (arr[i][j] != 1)
            cout << arr[i][j] - 1;
          else
            cout << 9;
        } else if (i == 2 && j == 2) {
          if (arr[i][j] != 1)
            cout << arr[i][j] - 1;
          else
            cout << 9;
        } else if (i == 3 && j == 6) {
          if (arr[i][j] != 1)
            cout << arr[i][j] - 1;
          else
            cout << 9;
        } else if (i == 4 && j == 9) {
          if (arr[i][j] != 1)
            cout << arr[i][j] - 1;
          else
            cout << 9;
        } else if (i == 5 && j == 3) {
          if (arr[i][j] != 1)
            cout << arr[i][j] - 1;
          else
            cout << 9;
        } else if (i == 6 && j == 5) {
          if (arr[i][j] != 1)
            cout << arr[i][j] - 1;
          else
            cout << 9;
        } else if (i == 7 && j == 8) {
          if (arr[i][j] != 1)
            cout << arr[i][j] - 1;
          else
            cout << 9;
        } else if (i == 8 && j == 1) {
          if (arr[i][j] != 1)
            cout << arr[i][j] - 1;
          else
            cout << 9;
        } else if (i == 9 && j == 4) {
          if (arr[i][j] != 1)
            cout << arr[i][j] - 1;
          else
            cout << 9;
        } else {
          cout << arr[i][j];
        }
      }
      cout << endl;
    }
  }
}