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>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")


    #define int long long
    

#define YES cout<<"YES\n";return; 
#define NO cout<<"NO\n";return; 
#define write(a) cout<<a<<"\n";return; 
#define F first 
#define S second

using namespace std;
using vi=vector<int>; 
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count()); //uid<int> u1(5, 10); u1(rnd);
long long powmod(long long a,long long b,long long mod) {
        if (b==0 || a==1)  {if (mod==1) return 0; else return 1; }
       
        if (b%2==0) { long long k=powmod(a,b/2,mod); return (k*k)%mod; }
        else {long long k=powmod(a,b/2,mod); return ( (k*k) %mod *a)% mod; }
}
long long gcd(long long a, long long b) {
    if (a==0) return b;
    if (b==0) return a;
    if (a>b) return gcd(a%b,b); else return gcd(b%a,a);
}
int prime(int p) { // 1 - простое
if (p==1) return 0; 
    for (int i=2;i*i<=p;i++) {
    if (p%i==0 && i<p) return i;  
    }
    return 1;
}
        
 int inv(int a, int mod) {
    return powmod(a,mod-2,mod); 
}
            
 int random_modul() {
        
    uniform_int_distribution <int> u1(1e9, 2e9);
    int pepega=u1(rnd); 
    while (prime(pepega)!=1) pepega=u1(rnd); 
    return pepega; 
}
    
template <typename T1> void ifmax(T1 &max, T1 kek) {
    if (kek>max) max=kek; 
    return; 
}
template <typename T1> void ifmin(T1 &min, T1 kek) {
    if (kek<min) min=kek;
    return; 
}
     

template <typename T1> istream& operator>> (std::istream& in, pair<T1,T1> & a) {
in>>a.first>>a.second; 
return in; 
    
}
    
template <typename T1> istream& operator>> (std::istream& in, vector<T1> & a) {
    for (int i=0;i<a.size();i++) in>>a[i]; 
    return in;
} 
     
template <typename T1>  ostream& operator<< (std::ostream& out, vector <T1> &b) {
    for (int i=0;i<b.size();i++) out<<b[i]; 
    return out; 
} 
template <typename T1> void sort(vector <T1> &m) {
    sort(m.begin(),m.end());
}

template <typename T1> void reverse(vector <T1> &m) {
 reverse(m.begin(),m.end());
}
int mod=1000000007; 
int fac[200005];
int inv_fac[200005]; 
int co(int a, int b) {
    int ans=1;
    ans*=fac[b]; 
    ans*=inv_fac[b-a]; ans%=mod; 
    ans*=inv_fac[a]; 
    return ans%mod; 
}    


void solve () {
           
           
     /* ---- !! F, S !!  ---- */
         
        int a;
        cin>>a;
        vector <vector<int>> m1(a, vector <int> (a));
        for (int i=0;i<a;i++) {
            for (int j=0;j<a;j++) {
                char c; cin>>c; m1[i][j]=(c-'0'); 
            }
        }
        vector <vector<int>> m2(a, vector <int> (a));
        for (int i=0;i<a;i++) {
            for (int j=0;j<a;j++) {
                char c; cin>>c; m2[i][j]=(c-'0'); 
            }
        }
        vector <int> lol1(a);
        vector <int> lol2(a); 
        for (int i=0;i<a;i++) {
            for (int j=0;j<a;j++) {
                if (m1[i][j]!=m2[i][j]) {
                   m1[i][j]=1; 
                } else m1[i][j]=0; 
            }
        }
        for (int i=0;i+1<a;i++) {
            for (int j=0;j+1<a;j++) {
                if ((m1[i][j]+m1[i+1][j]+m1[i+1][j+1]+m1[i][j+1])%2) {NO}
            }
        }
        {YES}
        
        
    
    /* --------- */  
    return;
    }
         
         
         
    signed main() {
   ios_base::sync_with_stdio(0);
   cin.tie(0);
   cout.tie(0); 
           
           
      cout<<fixed<<setprecision(10); 
           
           
        int multitest = 1 ;    
           
      int tututu;
      tututu=1;
      
      
     int combinatorics = 1;  
      
      if (combinatorics) { 
     fac[1]=1;
     fac[0]=1; 
     for (int i=2;i<200005;i++) {
         fac[i]=fac[i-1]*i; fac[i]%=mod; 
     }
     for (int i=0;i<200005;i++) {
         inv_fac[i]=inv(fac[i],mod); 
     }
      } 
      
      
    
      
      if (multitest) cin>>tututu; // если нет запросов, то закоментить
            
      for (int qwerty=0;qwerty<tututu;qwerty++) solve();
          
           
    return 0;
}
