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 inf = 0x3f3f3f3f;
const int maxn = 100000 + 10;
int n, m, ans;
char s[20][20];
void print(int a[], int n, int st = 0) {
  for (int i = 0; i < n; i++) {
    if (i != 0) printf(" ");
    printf("%d", a[i + st]);
  }
  printf("\n");
}
void init() {
  for (int i = 0; i < 9; i++) scanf("%s", s[i]);
}
void doit() {
  int a[3];
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
      for (int k = 0; k < 9; k++) {
        if (s[i * 3 + j][k] == '1') {
          a[j] = k;
          break;
        }
      }
    }
    for (int j = 0; j < 3; j++) s[i * 3 + j][a[(j + 1) % 3]] = '1';
  }
  for (int i = 0; i < 9; i++) printf("%s\n", s[i]);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    init();
    doit();
  }
  return 0;
}