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;
template <class T>
inline T bigMod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T modInverse(T a, T M) {
  return bigMod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  a = abs(a);
  b = abs(b);
  return (a / gcd(a, b)) * b;
}
template <class T>
inline T diss(T x1, T x2, T y1, T y2) {
  return sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)));
}
template <class T>
inline T disss(T x1, T x2, T z1, T y1, T y2, T z2) {
  return sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)) +
              ((z1 - z2) * (z1 - z2)));
}
template <class T>
inline string int2String(T a) {
  ostringstream str;
  str << a;
  return str.str();
}
template <class T>
inline T string2int(string a) {
  stringstream str(a);
  T b;
  str >> b;
  return b;
}
const int dr[] = {0, 1, 0, -1, -1, 1, 1, -1, -2, -2, 2, 2, -1, -1, 1, 1};
const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1, -1, 1, -1, 1, -2, 2, -2, 2};
int main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    int v[9][9];
    for (int i = 0; i < 9; i++) {
      string s;
      cin >> s;
      for (int j = 0; j < 9; j++) {
        v[i][j] = s[j] - '0';
      }
    }
    for (int i = 0; i < 9; i++) {
      for (int j = 0; j < 9; j++) {
        long long a = v[i][j];
        if (i == 0 && j == 0) {
          if (a == 9)
            v[i][j] = --a;
          else
            v[i][j] = ++a;
        } else if (i == 1 && j == 4) {
          if (a == 9)
            v[i][j] = --a;
          else
            v[i][j] = ++a;
        } else if (i == 2 && j == 8) {
          if (a == 9)
            v[i][j] = --a;
          else
            v[i][j] = ++a;
        } else if (i == 4 && j == 1) {
          if (a == 9)
            v[i][j] = --a;
          else
            v[i][j] = ++a;
        } else if (i == 5 && j == 5) {
          if (a == 9)
            v[i][j] = --a;
          else
            v[i][j] = ++a;
        } else if (i == 3 && j == 6) {
          if (a == 9)
            v[i][j] = --a;
          else
            v[i][j] = ++a;
        } else if (i == 8 && j == 2) {
          if (a == 9)
            v[i][j] = --a;
          else
            v[i][j] = ++a;
        } else if (i == 6 && j == 3) {
          if (a == 9)
            v[i][j] = --a;
          else
            v[i][j] = ++a;
        } else if (i == 7 && j == 7) {
          if (a == 9)
            v[i][j] = --a;
          else
            v[i][j] = ++a;
        }
      }
    }
    for (int i = 0; i < 9; i++) {
      for (int j = 0; j < 9; j++) {
        cout << v[i][j];
      }
      cout << endl;
    }
  }
  return 0;
}