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 ll          long long
#define pb          push_back
#define ppb         pop_back
#define eb          emplace_back
#define endl		'\n'
#define max(a,b)	((a)>(b)?(a):(b))
#define min(a,b)	((a)<(b)?(a):(b))
#define vi          vector<int>
#define vii         vector<pair<int, int> >
#define mp          make_pair
#define mem1(a)     memset((a),-1,sizeof(a))
#define mem0(a)     memset((a),0,sizeof(a))
#define memf(a)     memset((a),false,sizeof(a))
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define gcd(a,b)    __gcd((a),(b))
#define lcm(a,b)    (a)*((b)/gcd((a),(b)))
#define DECIMAL(n)  cout << fixed ; cout << setprecision(n);
#define FAST        ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define mi          map<int,int>
#define sz(x)       (ll)(x).size()
#define rep(i,a,b)  for(int i=(a);i<(b);i++)
#define pii         pair<int,int>

int solve()
{
	int n;
	cin >> n;
	string s[n];
	string a[n];
	rep(i, 0, n)
	cin >> s[i];
	rep(i, 0, n)
	cin >> a[i];
	rep(i, 0, n)
	{
		if (s[0][i] != a[0][i])
		{
			rep(j, 0, n)
			{
				s[j][i] = '1'^'0'^s[j][i];
			}
		}
	}
	rep(i, 1, n)
	{
		int r = 0, e = 0;
		rep(j, 0, n)
		{
			if (s[i][j] != a[i][j])
				r++;
			else
				e++;
		}
		if (!(e == n || r == n))
		{
			cout << "NO" << endl;
			return 0;
		}
	}
	cout << "YES" << endl;
	return 0;
}
signed main()
{
	FAST
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	int t = 1;
	cin >> t;
	while (t)
	{
		solve();
		t--;
	}
	return 0;
}