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 <class T1, class T2, class T3 = hash<T1>>
using umap = unordered_map<T1, T2, T3>;
template <class T>
using uset = unordered_set<T>;
template <class T>
using vec = vector<T>;
const long long infll = numeric_limits<long long>::max() >> 1;
const int inf = numeric_limits<int>::max() >> 1;
const int N = 9;
int tc;
int mat[N][N];
void input() {
  for (int i = 0; i < 9; ++i) {
    string t;
    cin >> t;
    for (int j = 0; j < 9; ++j) {
      mat[i][j] = t[j] - '0';
    }
  }
}
void solve() {
  for (int i = 0; i < 3; ++i) {
    for (int j = 0; j < 3; ++j) {
      int x = 3 * i + j;
      int y = 3 * j + i;
      mat[x][y] = mat[x][y] == 1 ? 2 : 1;
    }
  }
  for (int i = 0; i < 9; ++i) {
    for (int j = 0; j < 9; ++j) {
      cout << mat[i][j];
    }
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> tc;
  while (tc--) {
    input();
    solve();
  }
}