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 int MOD = 1000 * 1000 * 1000 + 7;
const long long int INF = 1ll * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 7;
const long long int MOD2 = 998244353;
const long double PI = 3.141592653589793;
long long int gcd(long long int a, long long int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    char arr[9][9];
    for (int i = 0; i < 9; i++) {
      for (int j = 0; j < 9; j++) {
        cin >> arr[i][j];
      }
    }
    arr[0][0] = (arr[0][0] + 1);
    if (arr[0][0] > '9') arr[0][0] = '1';
    arr[1][4] = (arr[1][4] + 1);
    if (arr[1][4] > '9') arr[1][4] = '1';
    arr[2][8] = (arr[2][8] + 1);
    if (arr[2][8] > '9') arr[2][8] = '1';
    arr[3][6] = (arr[3][6] + 1);
    if (arr[3][6] > '9') arr[3][6] = '1';
    arr[4][1] = (arr[4][1] + 1);
    if (arr[4][1] > '9') arr[4][1] = '1';
    arr[6][3] = (arr[6][3] + 1);
    if (arr[6][3] > '9') arr[6][3] = '1';
    arr[5][5] = (arr[5][5] + 1);
    if (arr[5][5] > '9') arr[5][5] = '1';
    arr[7][7] = (arr[7][7] + 1);
    if (arr[7][7] > '9') arr[7][7] = '1';
    arr[8][2] = (arr[8][2] + 1);
    if (arr[8][2] > '9') arr[8][2] = '1';
    for (int i = 0; i < 9; i++) {
      for (int j = 0; j < 9; j++) {
        cout << arr[i][j];
      }
      cout << "\n";
    }
  }
}