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.
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include <unordered_map>
//#include"testlib.h"
#define endl '\n'
#define all(v) v.begin(),v.end()
#define allr(s) s.rbegin(),s.rend()
#define RT(s) return cout<<s,0
#define sz(s)    (int)(s.size())
//#define PI acos(-1)
#define EPS 1e-8
#define watch(x) cout << (#x)<<" = "<<x<<endl
#define mk(x, y) make_pair(x, y)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
int dy[] = { 1, -1, 0, 0, -1, 1, 1, -1 };
int dx[] = { 0, 0, 1, -1, 1, -1, 1, -1 };
void file() {
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
#else
	//freopen("gcd.in", "r", stdin);
	//freopen("out.txt", "w", stdout);
#endif
}

void fast() {
	std::ios_base::sync_with_stdio(0);
	cin.tie(NULL);
}
string flib(const string& str) {
	string ret = "";
	for (auto& ch : str)
		ret.push_back(1 - (ch - '0') + '0');
	return ret;
}
vector<string> a, b;
int n;
bool test_case() {
	cin >> n;
	a = b = vector<string>(n);
	for (auto& it : a) cin >> it;
	for (auto& it : b) cin >> it;
	/*bool ok = true;
	for (int i = 0; i < n; i++)
		ok &= (a[i] == b[i] || flib(a[i]) == b[i]);
	if (ok) return true;*/
	vi row(n), col(n);
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			if (a[i][j] == b[i][j]) {
				row[i] = col[j] = true;
				continue;
			}
			if (row[i] && col[j]) return false;
			if (!row[i])
				a[i] = flib(a[i]);
			else {
				for (int k = 0; k < n; k++)
					a[k][j] = (1 - (a[k][j] - '0') + '0');
			}
			row[i] = col[j] = 1;
		}
	}
	return true;
}
int main() {
	file();
	fast();
	int t; cin >> t;
	while (t--) {
		cout << (test_case() ? "YES\n" : "NO\n");
	}
	return 0;
}