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;
long long modpow(long long a, long long b, long long m = 1000000007) {
  a %= m;
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long bpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long modinv(long long a) { return modpow(a, 1000000007 - 2, 1000000007); }
void graph(vector<vector<long long> > &adj, long long m) {
  long long x, y;
  for (long long i = 0; i < m; i++) {
    cin >> x >> y;
    adj[x - 1].push_back(y - 1);
    adj[y - 1].push_back(x - 1);
  }
}
const int mxN = 2e5 + 5;
void solve() {
  string arr[9];
  for (long long i = 0; i < 9; i++) cin >> arr[i];
  for (long long i = 0; i < 3; i++) {
    for (long long j = 0; j < 3; j++) {
      arr[3 * i + j][3 * j + i] =
          char(((arr[3 * i + j][3 * j + i] - '0') % 9 + 1) + '0');
    }
  }
  for (long long i = 0; i < 9; i++) cout << arr[i] << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}