// Problem: D. Anti-Sudoku
// Contest: Codeforces - Codeforces Round #634 (Div. 3)
// URL: https://codeforces.com/problemset/problem/1335/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 2022-03-02 13:54:04
//
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
ll ans = 1;
while (b) {
if (b & 1) ans = ans * a%mod;
a = a * a %mod;
b >>= 1;
}
return ans;
}
inline int read () {
int x = 0, f = 0;
char ch = getchar();
while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return f?-x:x;
}
template<typename T> void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) print(x/10);
putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
return (a + b) %mod;
}
inline ll inv (ll a) {
return qmi(a, mod - 2);
}


void solve() {
for (int i = 0; i < 9; i ++) {
string s;
cin >> s;
for (auto & t : s) if (t == '2') t = '1';
cout << s << endl;
}
}
int main () {
int t;
t =1;
cin >> t;
while (t --) solve();
return 0;
}


把所有2换成1就行了。因为本身给定的是一个解。满足原始数独的要求。把所有2换成 1,易得题目的要求