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:
def main():
    # from collections import defaultdict
    from collections import Counter
    #import math
    # from fractions import Fraction
    # import heapq
    # from bisect import bisect_left
    import sys
    input = sys.stdin.buffer.readline
    #mod=998244353
    def GCD(x, y):
        while (y):
            x, y = y, x % y
        return x
    def lcm(a,b):
        return (a*b)//GCD(a,b)
    # Modified merge for inversion
    def merge(a,start,mid,end,d):
        p=start
        q=mid+1
        arr=[0]*(end-start+1)
        k=0
        coun=0
        for i in range(start,end+1):
            if p > mid:
                arr[k]=a[q]
                q+=1
                k+=1
            elif q > end:
                arr[k]=a[p]
                d[a[p]] += coun
                k+=1
                p+=1
            elif a[p] < a[q]:
                arr[k]=a[p]
                d[a[p]]+=coun
                k+=1
                p+=1
            else:
                arr[k]=a[q]
                k+=1
                q+=1
                coun+=1

        for i in range(k):
            a[start]=arr[i]
            start+=1
    def mergesort(a, start, end,d):
        if start<end:
            mid=(start+end)//2
            mergesort(a,start,mid,d)
            mergesort(a,mid+1,end,d)
            merge(a,start,mid,end,d)
    # inversion return dic for every index inversion count
    def inversion(l):
        d={}
        for i in l:
            d[i]=0
        mergesort(l,0,len(l)-1,d)
        return d
    def binary_search_lower(A,low,high,k):
        if high-low!=1:
            mid=(low+high)//2
            x = A[mid]
            if x>=k:
                return binary_search_lower(A,low,mid,k)
            else:
                return binary_search_lower(A,mid,high,k)
        if A[low]>=k:
            return low
        else:
            return high
    # Function to find power
    def binary_expo(x,n):
        if n==0:
            return 1
        elif n%2==0:
            return binary_expo(x*x,n//2)
        else:
            return x*binary_expo(x*x,(n-1)//2)
    # Function to find power with given modulus
    def modu_expo(x,n,m):
        if n==0:
            return 1
        elif n%2==0:
            return modu_expo((x*x)%m,n//2,m)
        else:
            return (x*modu_expo((x*x)%m,(n-1)//2,m))%m
    # Function for modular multiplicative inverse via using extended_GCD_Euclid's
    def eEuclid(c, m):
        if m == 0:
            global d
            global x
            global y
            d = c
            x = 1
            y = 0
        else:
            eEuclid(m, c % m)
            temp = x
            x = y
            y = temp - (c // m) * y
    def mod_Inv(c, m):
        eEuclid(c, m)
        return (x % m + m) % m
    # Function for nCr%p using fermat (p is prime)
    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 change(x):
        if x=="9":
            return "1"
        else:
            return "9"

    #*************************************************************
    for _ in range(int(input())):
        l=[]
        for i in range(9):
            l.append(list(str(input())))
        for i in l:
            i.pop()
            i.pop()
            i.pop()
            i.pop(0)
            i.pop(0)
        x=l[0][0]
        l[0][0]=change(x)
        x = l[1][4]
        l[1][4] = change(x)
        x = l[2][8]
        l[2][8] = change(x)
        x = l[3][2]
        l[3][2] = change(x)
        x = l[4][3]
        l[4][3] = change(x)
        x = l[5][7]
        l[5][7] = change(x)
        x = l[6][1]
        l[6][1] = change(x)
        x = l[7][5]
        l[7][5] = change(x)
        x = l[8][6]
        l[8][6] = change(x)
        #print(l)
        for i in range(9):
            for j in range(9):
                print(l[i][j],end="")
            print()

if __name__ == "__main__":
    main()