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;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MX = 2e5 + 7;
const int mod = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    k >>= 1;
  }
  return res;
}
int mp[10][10];
void f(int x, int y) {
  int num = mp[x][y];
  for (int i = 1; i < 10; ++i) {
    if (i != num) {
      mp[x][y] = i;
      return;
    }
  }
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    for (int i = 1; i <= 9; ++i) {
      for (int j = 1; j <= 9; ++j) {
        scanf("%1d", &mp[i][j]);
      }
    }
    f(1, 1);
    f(2, 5);
    f(3, 9);
    f(4, 2);
    f(5, 4);
    f(6, 7);
    f(7, 3);
    f(8, 6);
    f(9, 8);
    for (int i = 1; i <= 9; ++i) {
      for (int j = 1; j <= 9; ++j) {
        cout << mp[i][j];
      }
      cout << '\n';
    }
  }
  return 0;
}