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;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long power(long long a, long long n) {
  if (n == 0) return 1;
  long long p = power(a, n / 2);
  p = (p * p);
  if (n % 2)
    return (p * a);
  else
    return p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    vector<string> v(9);
    for (int i = 0; i < 9; i++) {
      cin >> v[i];
    }
    if (v[0][0] == '1')
      v[0][0] = '2';
    else
      v[0][0] = '1';
    if (v[1][3] == '1')
      v[1][3] = '2';
    else
      v[1][3] = '1';
    if (v[2][6] == '1')
      v[2][6] = '2';
    else
      v[2][6] = '1';
    if (v[3][1] == '1')
      v[3][1] = '2';
    else
      v[3][1] = '1';
    if (v[4][4] == '1')
      v[4][4] = '2';
    else
      v[4][4] = '1';
    if (v[5][7] == '1')
      v[5][7] = '2';
    else
      v[5][7] = '1';
    if (v[6][2] == '1')
      v[6][2] = '2';
    else
      v[6][2] = '1';
    if (v[7][5] == '1')
      v[7][5] = '2';
    else
      v[7][5] = '1';
    if (v[8][8] == '1')
      v[8][8] = '2';
    else
      v[8][8] = '1';
    for (int i = 0; i < 9; i++) cout << v[i] << endl;
    cout << endl;
  }
  return 0;
}