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 = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
const int N = (int)1e5 + 10;
void TimeExecution() {
  cout << "\n\n===>> Time Execution: " << clock() / (double)1000 << " sec(s).";
}
int t;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    string s[10];
    for (int i = (1); i <= (9); i++) cin >> s[i];
    s[1][0] = s[1][1];
    s[2][3] = s[2][4];
    s[3][6] = s[3][7];
    s[4][1] = s[4][2];
    s[5][4] = s[5][5];
    s[6][7] = s[6][8];
    s[7][2] = s[7][3];
    s[8][5] = s[8][6];
    s[9][8] = s[9][7];
    for (int i = (1); i <= (9); i++) cout << s[i] << endl;
  }
  return 0;
}