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>
long long exp(long long x, long long y, long long p) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res * x % p) % p;
    x = (x * x) % p;
    y /= 2;
  }
  return res;
}
long long expo(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res * x % ((int)1e9 + 7)) % ((int)1e9 + 7);
    x = (x * x) % ((int)1e9 + 7);
    y /= 2;
  }
  return res;
}
long long add(long long a, long long b) {
  return (a % ((int)1e9 + 7) + b % ((int)1e9 + 7) + ((int)1e9 + 7)) %
         ((int)1e9 + 7);
}
long long sub(long long a, long long b) {
  return (a % ((int)1e9 + 7) - b % ((int)1e9 + 7) + ((int)1e9 + 7)) %
         ((int)1e9 + 7);
}
long long mul(long long a, long long b) {
  return ((a % ((int)1e9 + 7)) * (b % ((int)1e9 + 7)) + ((int)1e9 + 7)) %
         ((int)1e9 + 7);
}
long long inv(long long x) { return expo(x, ((int)1e9 + 7) - 2); }
using namespace std;
int sv[2000006] = {1};
void sve() {
  for (int i = 2; i * i < 100001; i++) {
    if (!sv[i]) {
      for (int j = i * i; j < ((int)2e6 + 6); j += i) sv[j] = 1;
    }
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n = 9;
    vector<string> v;
    for (int i = 0; i < n; i++) {
      string a;
      cin >> a;
      v.push_back(a);
    }
    v[0][0] = '0' + (v[0][0] % 9) + 1;
    v[1][3] = '0' + (v[1][3] % 9) + 1;
    v[2][6] = '0' + (v[2][6] % 9) + 1;
    v[3][1] = '0' + (v[3][1] % 9) + 1;
    v[4][4] = '0' + (v[4][4] % 9) + 1;
    v[5][7] = '0' + (v[5][7] % 9) + 1;
    v[6][2] = '0' + (v[6][2] % 9) + 1;
    v[7][5] = '0' + (v[7][5] % 9) + 1;
    v[8][8] = '0' + (v[8][8] % 9) + 1;
    for (int i = 0; i < n; i++) {
      cout << v[i] << endl;
    }
  }
}