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.
a=[]
acp=[]
b=[]
n = 0

def printt():
    for i in range(n):
        print(a[i])
    print("printed.")
    
def solveRow():
    for i in range(n):
        if acp[0][i] != b[0][i]:
            for j in range(n):
                if acp[j][i] == '0':
                    acp[j][i] = '1'
                else:
                    acp[j][i] = '0'
    #printt()
    for i in range(1,n):
        match = 0
        for j in range(n):
            if acp[i][j] != b[i][j]:
                match += 1
        if match != 0 and match != n:
            return False
    return True

def copy():
    global acp
    acp=[]
    for i in range(n):
        acp.append(a[i])
        
def solve():
    global n
    n=int(input())
    global a
    a=[]
    for i in range(n):
        s=input()
        a.append(list(s))
    global b
    b=[]
    input()
    for i in range(n):
        s=input()
        b.append(list(s))
    copy()
    if solveRow() == True:
        print("YES")
        return
    for i in range(n):
        if a[0][i] == '0':
            a[0][i] = '1'
        else:
            a[0][i] = '0'
    copy()
    if solveRow() == True:
        print("YES")
    else:
        print("NO")
    
    
def main():
    t=int(input())
    for i in range(t):
        solve()

if __name__ == "__main__":
    main()
