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);
vector<bool> prime;
void primes(long long n) {
  prime.resize(n + 1);
  prime.assign(n + 1, true);
  prime[0] = 0;
  prime[1] = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (prime[i]) {
      for (long long j = (i * i); j < n; j = j + i) prime[j] = false;
    }
  }
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long powM(long long x, long long y, long long m) {
  long long ans = 1, r = 1;
  x = x % m;
  while (r > 0 && r <= y) {
    if (r & y) {
      ans = ans * x;
      ans = ans % m;
    }
    r = r << 1;
    x = x * x;
    x = x % m;
  }
  return ans;
}
long long powN(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long u = powN(a, n / 2);
  u = u * u;
  if (n % 2) u = u * a;
  return u;
}
long long modinv(long long a, long long m) { return powM(a, m - 2, m); }
long long ncr(long long n, long long r, long long m) {
  if (n < r) return 0;
  if (n == r || r == 0) return 1;
  if (r == 1) return n;
  if (n - r < r) return ncr(n, n - r, m);
  long long a = n;
  long long b = r;
  for (long long i = 1; i < r; i++) {
    a = (a * (n - i)) % m;
    b = (b * (r - i)) % m;
  }
  b = modinv(b, m);
  a = (a * b) % m;
  return a;
}
long long countbits(long long n) {
  long long cot = 0;
  while (n) {
    cot++;
    n >>= 1;
  }
  return cot;
}
long long check_prime(long long n) {
  long long ans = 1;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ans = 0;
      break;
    }
  }
  return ans;
}
void test_case() {
  long long n = 9;
  char a[n][n];
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      cin >> a[i][j];
    }
  }
  a[0][0] = a[0][1];
  a[1][3] = a[1][4];
  a[2][6] = a[2][7];
  a[3][1] = a[3][2];
  a[4][4] = a[4][5];
  a[5][7] = a[5][8];
  a[6][2] = a[6][3];
  a[7][5] = a[7][6];
  a[8][8] = a[8][7];
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      cout << a[i][j];
    }
    cout << "\n";
  }
  cout << "\n";
}
int main() {
  fast();
  long long T = 1;
  cin >> T;
  while (T--) {
    test_case();
  }
  return 0;
}