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.
#include <bits/stdc++.h>
#define ll long long
#define real long double
#define fr(i,l,n) for(ll i=l;i<n;i++)
#define br cout<<"\n"
#define al(fu) (fu).begin(), (fu).end()
#define prl(c) cout<<(c)<<"\n"
#define prv(v) {fr(qz,0,v.size()) cout<<(v)[qz]<<' '; br;}
#define alr(fu) (fu).rbegin(), (fu).rend()
#define mod 1000000007LL
#define mod1 998244353LL
#define nfi 1000000000000000013LL
#define skoree ios::sync_with_stdio(0); cin.tie(0)
#define danet(b) cout<<(b?"YES":"NO")
#define Danet(b) cout<<(b?"Yes":"No")
using namespace std;

char aq(char a){
    if(a=='0') return '1';
    else return '0';
}

int main(){
    skoree;
    ll o0o;
    cin>>o0o;
    while(o0o--){
        ll n;
        cin>>n;
        vector<string> a(n),b(n);
        fr(i,0,n) cin>>a[i];
        fr(i,0,n) cin>>b[i];
        fr(i,0,n) if(a[0][i]!=b[0][i]){
            fr(j,0,n) a[j][i]=aq(a[j][i]);
        }
        bool bb=1;
        fr(i,1,n) if(a[i]!=b[i]){
            fr(j,0,n) a[i][j]=aq(a[i][j]);
            if(a[i]!=b[i]) bb=0;
        }
        danet(bb);br;
    }
    return 0;
}
/*
*/
