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>
using namespace std;

#define pii             pair<int,int>
#define vi              vector<int>
#define mii             map<int,int>
#define pqs             priority_queue<int,vi,greater<int> >
#define mod             1000000007
#define int             long long
#define pb              push_back
#define inf             1e18
#define sp(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define w(t)            int t; cin>>t; while(t--)
#define mp              make_pair
#define endl            "\n"
#define pqb             priority_queue<int>
#define ceil(a , b)     ((a / b) + (a%b && 1))
#define rep(i,a,b)      for(int i=a;i<(int)b;i++)
#define repr(i,a,b)     for(int i=a;i>=(int)b;i--)
#define fastio          ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define all(v)          v.begin(), v.end()
#define ff              first
#define ss              second
template<typename... T> void pn(T... args) { ((cout << args << " "), ...); cout<<"\n"; }
template<typename... T> void ps(T... args) { ((cout << args << " "), ...); cout<<""; }
// void start()
// {
//     ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
// #ifndef ONLINE_JUDGE
//     freopen("input.txt", "r", stdin);
//     freopen("output.txt", "w", stdout);
// #endif
// }
int power(int a, int b,int P=mod) {
    int res = 1;
    for (; b; b >>= 1, a = 1ll * a * a % P)
        if (b & 1)
            res = 1ll * res * a % P;
    return res;
}

void solve()
{
    int n;
    cin>>n;
    vector<string> a(n),b(n);
    for(int i=0; i<n; i++)  cin>>a[i];
    for(int i=0; i<n; i++)  cin>>b[i];
    int arr[n][n];
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++)
        {
            if(a[i][j]==b[i][j])    arr[i][j]=0;
            else    arr[i][j]=1;
        }
    }
    for(int i=0; i<n; i++)
    {
        if(arr[0][i]==1)
        {
            for(int j=0; j<n; j++)
            {
                arr[j][i]^=1;
            }
        }
    }

    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++)
        {
            if(arr[i][j]!=arr[i][0])
            {
                cout<<"NO"<<endl;
                return;
            }
        }
    }
    cout<<"YES"<<endl;
}
int32_t main()
{
    fastio;
    w(t)
    {
        solve();
    }

    return 0;
}