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 prime[1000002];
using namespace std;
int Prime(int a) {
  for (int i = 2; i <= a / 2; ++i) {
    if (a % i == 0) {
      return 0;
    }
  }
  return 1;
}
void sieve() {
  for (int i = 0; i <= 1000001; i++) prime[i] = 0;
  int maxn = 1000001;
  for (int i = 2; i <= maxn; i++) {
    if (prime[i] == 0) {
      for (int j = i; j <= maxn; j += i) {
        prime[j]++;
      }
    }
  }
}
void fibo(long long a[], long long n) {
  long long x = 0;
  long long y = 1;
  a[0] = 0;
  a[1] = 1;
  for (long long i = 2; i < n; ++i) {
    a[i] = (x + y) % 10;
    x = y;
    y = a[i];
  }
}
long long f(long long w[], long long a[][2], long long i, long long y) {
  for (long long j = 0; j < y; ++j) {
    if (a[i][2] <= w[j]) {
      cout << a[i][2];
      return w[j];
    }
  }
  return w[y - 1];
}
long long s(long long v[], long long a[][2], long long i, long long x) {
  for (long long j = 0; j < x; ++j) {
    if (a[i][1] >= v[j]) {
      cout << " " << a[i][1] << " ";
      return v[j];
    }
  }
  return v[0];
}
bool binarySearch(vector<long long> arr, int l, int r, int x) {
  if (r >= l) {
    int mid = l + (r - l) / 2;
    if (arr[mid] == x) return 1;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return 0;
}
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  char c;
  long long i, j = 0, t;
  cin >> t;
  while (t--) {
    for (i = 0; i < 9; ++i) {
      for (j = 0; j < 9; ++j) {
        cin >> c;
        if (c == '2') {
          cout << 1;
        } else {
          cout << c;
        }
      }
      cout << endl;
    }
    cout << endl;
  }
  return 0;
}