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 PYTHON3):

Solution:
from sys import stdin
from collections import deque
mod = 10**9 + 7
import sys
import random
# sys.setrecursionlimit(10**6)
from queue import PriorityQueue
# def rl():
#     return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
from itertools import permutations
from math import factorial as f

# def ncr(x, y):
#     return f(x) // (f(y) * f(x - y))
def ncr(n, r, p):
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den,
                      p - 2, p)) % p

def solve(n):
    return (n*(n+1))//2

# v1 ko root manio

for _ in range(10**5):
    pass
for _ in range(10**5):
    pass
for _ in range(10**5):
    pass
t = int(input())


for i in range(t):


    # n = int(input())

    l = []

    for i in range(9):
        laa = list(input())
        l.append(laa)

    cnt = 0
    for i in range(9):
        za = l[i][0]
        if i == 0:
            l[i][3] = za
        if i == 1:
            l[i][6] = za
        if i == 2:
            l[i][2] = za

        if i == 3:
            l[i][4] = za
        if i == 4:
            l[i][7] = za

        if i == 5:
            l[i][1] = za

        if i == 6:
            l[i][5] = za
        if i == 7:
            l[i][8] = za

        if i == 8:
            l[i][0] = l[0][0]

    for i in l:
        print(''.join(i))