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 long long inf = 0x3f3f3f3f;
const int maxn = 2e5 + 7;
const long long mod = 1e9 + 7;
inline int read() {
  int x = 0, f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
char a[15][15];
int main() {
  long long n, b;
  long long t;
  cin >> t;
  while (t--) {
    for (int i = 0; i < 9; i++) cin >> a[i];
    a[0][0] = a[0][1];
    a[1][3] = a[1][4];
    a[2][6] = a[2][7];
    a[3][1] = a[3][2];
    a[4][4] = a[4][5];
    a[5][7] = a[5][8];
    a[6][2] = a[5][2];
    a[7][5] = a[6][5];
    a[8][8] = a[7][8];
    for (int i = 0; i < 9; i++) cout << a[i] << endl;
  }
}