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>
#pragma GCC optimize("Ofast")
using namespace std;
using ll = long long;
using datas = pair<ll, ll>;
using ddatas = pair<long double, long double>;
using tdata = pair<ll, datas>;
using vec = vector<ll>;
using mat = vector<vec>;
using pvec = vector<datas>;
using pmat = vector<pvec>;
const ll mod = 1000000007;
const ll inf = 1LL << 60;
const long double PI = acos(-1);
const long double eps = 1e-9;
template <class T>
inline bool chmax(T& a, T b) {
  bool x = a < b;
  if (x) a = b;
  return x;
}
template <class T>
inline bool chmin(T& a, T b) {
  bool x = a > b;
  if (x) a = b;
  return x;
}
void startupcpp() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}
long double distance(ddatas x, ddatas y) {
  long double a = x.first - y.first, b = x.second - y.second;
  return sqrt(a * a + b * b);
}
ll modinv(ll a) {
  ll b = mod, u = 1, v = 0, t;
  while (b) {
    t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  return (u + mod) % mod;
}
ll moddevide(ll a, ll b) { return (a * modinv(b)) % mod; }
vec modncrlistp, modncrlistm;
ll modncr(ll n, ll r) {
  if (n < r) return 0;
  ll i, size = modncrlistp.size();
  if (size <= n) {
    modncrlistp.resize(n + 1);
    modncrlistm.resize(n + 1);
    if (!size) {
      modncrlistp[0] = modncrlistm[0] = 1;
      size++;
    }
    for (i = size; i < (ll)n + 1; ++i) {
      modncrlistp[i] = modncrlistp[i - 1] * i % mod;
      modncrlistm[i] = modinv(modncrlistp[i]);
    }
  }
  return modncrlistp[n] * modncrlistm[r] % mod * modncrlistm[n - r] % mod;
}
ll modpow(ll a, ll n) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
ll gcd(ll a, ll b) {
  if (!b) return a;
  return (a % b == 0) ? b : gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll countdigits(ll n) {
  ll ans = 0;
  while (n) {
    n /= 10;
    ans++;
  }
  return ans;
}
ll sumdigits(ll n) {
  ll ans = 0;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
int main() {
  startupcpp();
  int codeforces;
  cin >> codeforces;
  while (codeforces--) {
    ll i, j;
    string s;
    for (i = 0; i < (ll)9; ++i) {
      cin >> s;
      for (j = 0; j < (ll)9; ++j)
        if (s[j] == '1') s[j] = '2';
      cout << s << "\n";
    }
  }
  return 0;
}