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 pi 3.1415926535
#define int long long
#define endl "\n"
#define yes cout << "YES\n"
#define no cout << "NO\n"
#define ll long long
#define setbits(x) __builtin_popcountll(x)
#define pb push_back
#define sz(x) (int)x.size()
#define f(i,a,n) for(int i = a ; i < n ; i++)
#define F first
#define S second
#define all(c) (c).begin(),(c).end()
#define get(a,n) f(i,0,n) cin >> a[i];
#define show(a,n) f(i,0,n) cout << a[i] << " "; cout << endl;
#define nitro ios:: sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define trace(x) cerr << '>' << #x << ':' << (x) << "\n"
#define trace2(x,y) cerr<< '>' << #x << ':' << (x) << " | " << #y << ':' << (y) << "\n"
#define trace3(a,b,c) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<"\n"
#define trace4(a,b,c,d) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<"\n"

typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef map<int,int> mii;
typedef map<ll,ll> mll;

const int inf = 1e9;
const int inf64 = 1e18;
const int mod = inf + 7;
const int N = 2e5 + 50;

int n;

void flipRow(vector<vector<int> > &a, int r)
{
    for(int i = 0; i < n; i++)
        a[r][i] ^= 1;
}

void FlipCol(vector<vector<int> > &a, int c)
{
    for(int i = 0; i < n; i++)
        a[i][c] ^= 1;
}

void solve()
{
    cin >> n;
    vector<vector<int> > a(n, vector<int> (n, 0));
    vector<vector<int> > b(n, vector<int> (n, 0));

    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
            char c;
            cin >> c;
            a[i][j] = c - '0';
        }
    }
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
            char c;
            cin >> c;
            b[i][j] = c - '0';
        }
    }
    /**Just reform the first row and the first column(if needed) and then verify if the two matrices match**/
    if(a[0][0] != b[0][0]) flipRow(a, 0);
    for(int i = 1; i < n; i++)
    {
        if(a[0][i] != b[0][i])
            FlipCol(a, i);
    }
    for(int i = 1; i < n; i++)
    {
        if(a[i][0] != b[i][0])
            flipRow(a, i);
    }
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
            if(a[i][j] != b[i][j])
            {
                cout << "NO\n";
                return;
            }
        }
    }
    cout << "YES\n";
}

int32_t main()
{
    nitro;
    int tc = 1;
    cin >> tc;
    while(tc--) solve();
}
