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;
template <typename T>
T binpow(T n, T s, int base) {
  if (s <= 0) return 1LL;
  if (s % 2 == 0) {
    T b = binpow(n, s / 2, base);
    return (1LL * b * b) % base;
  } else
    return (1LL * binpow(n, s - 1, base) * n) % base;
}
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  if (b != 0) return gcd(b, a % b);
  return a;
}
void go(vector<string>& arr, int a, int b) {
  if (arr[a][b] == '9')
    arr[a][b] = '1';
  else
    arr[a][b] = arr[a][b] + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    vector<string> arr(9);
    for (int i = 0; i < 9; ++i) cin >> arr[i];
    go(arr, 0, 0);
    go(arr, 3, 1);
    go(arr, 6, 2);
    go(arr, 1, 3);
    go(arr, 4, 4);
    go(arr, 7, 5);
    go(arr, 2, 6);
    go(arr, 5, 7);
    go(arr, 8, 8);
    for (int i = 0; i < 9; ++i) {
      for (auto& j : arr[i]) cout << j;
      cout << endl;
    }
  }
  return 0;
}