1475_F. Unusual Matrix
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.
```python
def solve():
    for _ in range(int(input())):
        n=int(input())
        a=[]
        b=[]
        aux=[]
        for i in range(n):
            s=input()
            t=[]
            for j in s:
                t.append(int(j))
            a.append(t)
            aux.append(t)
        s=input()
        for i in range(n):
            s=input()
            t=[]
            for j in s:
                t.append(int(j))
            b.append(t)
        r=[0]*n
        c=[0]*n
        for j in range(n):
            if aux[0][j]!=b[0][j]:
                c[j]=1
        for i in range(n):
            for j in range(n):
                if c[j]==1:
                    aux[i][j]^=1
        f=0
        for i in range(n):
            change=0
            for j in range(n):
                if aux[i][j]!=b[i][j]:
                    change=1
            if change:
                for j in range(n):
                    aux[i][j]^=1
        if aux==b:
            print('YES')
            continue
        aux=[]
        for i in range(n):
            t=[]
            for j in range(n):
                t.append(a[i][j])
            aux.append(t)

        for j in range(n):
            aux[0][j] ^= 1
        c = [0] * n
        for j in range(n):
            if aux[0][j] != b[0][j]:
                c[j] = 1
        for i in range(n):
            for j in range(n):
                if c[j] == 1:
                    aux[i][j] ^= 1
        f = 0
        for i in range(n):
            change = 0
            for j in range(n):
                if aux[i][j] != b[i][j]:
                    change = 1
            if change:
                for j in range(n):
                    aux[i][j] ^= 1
        if aux == b:
            print('YES')
            continue
        print('NO')

if __name__ == '__main__':
    solve()
```