You are given two binary square matrices a and b of size n × n. A matrix is called binary if each of its elements is equal to 0 or 1. You can do the following operations on the matrix a arbitrary number of times (0 or more): 

  * vertical xor. You choose the number j (1 ≤ j ≤ n) and for all i (1 ≤ i ≤ n) do the following: a_{i, j} := a_{i, j} ⊕ 1 (⊕ — is the operation [xor](https://en.wikipedia.org/wiki/Exclusive_or) (exclusive or)). 
  * horizontal xor. You choose the number i (1 ≤ i ≤ n) and for all j (1 ≤ j ≤ n) do the following: a_{i, j} := a_{i, j} ⊕ 1. 



Note that the elements of the a matrix change after each operation.

For example, if n=3 and the matrix a is: $$$ \begin{pmatrix} 1 & 1 & 0 \\\ 0 & 0 & 1 \\\ 1 & 1 & 0 \end{pmatrix} $$$ Then the following sequence of operations shows an example of transformations: 

  * vertical xor, j=1. $$$ a= \begin{pmatrix} 0 & 1 & 0 \\\ 1 & 0 & 1 \\\ 0 & 1 & 0 \end{pmatrix} $$$ 
  * horizontal xor, i=2. $$$ a= \begin{pmatrix} 0 & 1 & 0 \\\ 0 & 1 & 0 \\\ 0 & 1 & 0 \end{pmatrix} $$$ 
  * vertical xor, j=2. $$$ a= \begin{pmatrix} 0 & 0 & 0 \\\ 0 & 0 & 0 \\\ 0 & 0 & 0 \end{pmatrix} $$$ 



Check if there is a sequence of operations such that the matrix a becomes equal to the matrix b.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 1000) — the size of the matrices.

The following n lines contain strings of length n, consisting of the characters '0' and '1' — the description of the matrix a.

An empty line follows.

The following n lines contain strings of length n, consisting of the characters '0' and '1' — the description of the matrix b.

It is guaranteed that the sum of n over all test cases does not exceed 1000.

Output

For each test case, output on a separate line: 

  * "YES", there is such a sequence of operations that the matrix a becomes equal to the matrix b; 
  * "NO" otherwise. 



You can output "YES" and "NO" in any case (for example, the strings yEs, yes, Yes and YES will be recognized as positive).

Example

Input


3
3
110
001
110

000
000
000
3
101
010
101

010
101
010
2
01
11

10
10


Output


YES
YES
NO

Note

The first test case is explained in the statements.

In the second test case, the following sequence of operations is suitable: 

  * horizontal xor, i=1; 
  * horizontal xor, i=2; 
  * horizontal xor, i=3; 



It can be proved that there is no sequence of operations in the third test case so that the matrix a becomes equal to the matrix b.
# Uses python3
import sys
import math

def xor(i):
    if i==0:
        i=1
    else:
        i=0
    return i
    
    
def fn(n,a,b):
    x=[]
    y=[]
    for i in range(n):
        a[i] = list(map(int, a[i]))
        b[i] = list(map(int, b[i]))
    if a==b:
        print("YES")
        return 0 
        
    for j in range(n):
        if int(a[0][j])!=int(b[0][j]):
            y.append(j)
    for j in y:
        for i in range(n):
            if int(a[i][j])==1:
               a[i][j]=0
            elif int(a[i][j])==0:
                a[i][j]=1
    if a==b:
        print("YES")
        return 0 
    else:
        for i in range(n):
            if a[i]!=b[i]:
                x.append(i)
        for i in x:
            for j in range(n):
                if int(a[i][j])==0:
                    a[i][j] = 1
                elif int(a[i][j])==1:
                    a[i][j]= 0
                    
        if a==b:
            print("YES")
            return 0 
        else:
            print("NO")
            return 0
    
if __name__ == '__main__':    
    input = sys.stdin.read()
    data = list(map(str, input.split()))
    t = int(data[0])
    l=0
    for i in range(t):
        n = int(data[l+1])
        a = [0]*n
        b = [0]*n
        for i in range(n):
            a[i] = list(data[i+l+2])
            b[i] = list(data[i+l+n+2])
        l=l+2*n+1
        fn(n,a,b)
