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 <typename _t>
inline void read(_t &x) {
  x = 0;
  _t fu = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') fu = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch & 15);
    ch = getchar();
  }
  x *= fu;
}
long long gcd(long long x, long long y) { return !x ? y : gcd(y % x, x); }
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, y, x);
  y -= a / b * x;
}
struct Mart {
  int a, b, c, d;
  Mart operator*(Mart &m) {
    Mart temp;
    temp.a = a * m.a + b * m.c;
    temp.b = a * m.b + b * m.d;
    temp.c = c * m.a + d * m.c;
    temp.d = c * m.b + d * m.d;
    return temp;
  }
  Mart() { a = b = c = d = 0; }
  Mart(int x[4]) {
    a = x[0];
    b = x[1];
    c = x[2];
    d = x[3];
  }
  Mart(int u, int v, int w, int x) {
    a = u;
    b = v;
    c = w;
    d = x;
  }
  int det() { return a * d - b * c; }
};
char s[15][15];
void change(int i, int j) {
  int now = s[i][j] - '0';
  if (now == 9)
    now = 1;
  else
    now = now + 1;
  s[i][j] = now + '0';
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    for (int i = 0; i < 9; i++) scanf("%s", s[i]);
    change(0, 0);
    change(1, 3);
    change(2, 6);
    change(3, 4);
    change(4, 7);
    change(5, 1);
    change(6, 8);
    change(7, 2);
    change(8, 5);
    for (int i = 0; i < 9; i++) printf("%s\n", s[i]);
  }
  return 0;
}