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>
void rd(T &...args) {
  ((cin >> args), ...);
}
template <typename... T>
void wr(T &...args) {
  ((cout << args << " "), ...);
}
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
void scan(int &x);
long long mult(long long a, long long b, long long mod);
long long binpow(long long a, long long b, long long mod);
template <typename T>
T lcm(T a, T b) {
  return (a * b) / gcd(a, b);
}
long long _sum(vector<int> a);
template <typename T>
void swap(T &n1, T &n2) {
  T temp;
  temp = n1;
  n1 = n2;
  n2 = temp;
}
void Amritesh() {
  vector<string> v;
  string s;
  for (int(i) = (0); (i) < (9); (i)++) {
    cin >> s;
    v.push_back(s);
  }
  for (int(i) = (0); (i) < (9); (i)++) {
    for (int(j) = (0); (j) < (9); (j)++) {
      if (v[i][j] == '1') {
        cout << "9";
      } else {
        cout << v[i][j];
      }
    }
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    Amritesh();
  }
  return 0;
}
long long _sum(vector<int> a) { return accumulate(a.begin(), a.end(), 0ll); }
long long mult(long long a, long long b, long long mod) {
  return (a * b) % mod;
}
long long binpow(long long a, long long b, long long mod) {
  long long powans = 1;
  for (; b > 0; a = mult(a, a, mod), b /= 2)
    if (b % 2 == 1) powans = mult(powans, a, mod);
  return powans;
}
void scan(int &x) {
  bool neg = false;
  register int c;
  x = 0;
  c = getchar();
  if (c == '-') {
    neg = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) x = (x << 1) + (x << 3) + c - 48;
  if (neg) x *= -1;
}