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.
//Classic
//g++ -std=c++17 -O2 -Wall a.cpp -o test
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define vll vector<ll>
#define vi vector<int>
#define vb vector<bool>
#define pi pair<int,int>
#define pll pair<ll,ll>
#define vp vector<pi>
#define vpll vector<pll>
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define F first
#define S second
#define For(i,a,b) for(ll i=a;i<b;i++)
//#define endl "\n"
#define debug2(x,y) cout<<"This side ----> "<<#x<<" -> "<<x<<" | "<<#y<<" -> "<<y<<endl;
#define debug(x) cout<<"This side    ----> "<<#x<<" -> "<<x<<endl
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(), x.rend()
#define mint map<int,int>
#define ciN cin
#define tu(a,s) get<a>(s)
#define tin tuple<ll,ll,ll>
#define ter(x,y,z) ((x)?y:z)
bool isGoogles = 0;
const ll maxn = 2e5 + 10;
const ll mod = 998244353;
const ll bits = 10;
ll caseNumber = 1;
////////////////////////////////////////////////////////////////

void jabru() {
	int n; cin >> n;
	int a[n][n], b[n][n];
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++) {
			char aa;
			cin >> aa;
			if (aa == '1')a[i][j] = 1;
			else a[i][j] = 0;
		}
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++) {
			char aa;
			cin >> aa;
			if (aa == '1')b[i][j] = 1;
			else b[i][j] = 0;
		}
	// don't flip first row
	vi col;
	for (int i = 0; i < n; i++) {
		if (a[0][i] != b[0][i]) {
			col.pb(i);
		}
	}
	for (auto x : col) {
		for (int i = 0; i < n; i++) {
			a[i][x] = !a[i][x];
		}
	}
	bool flag = 1;
	for (int i = 0; i < n; i++) {
		string forb = "";
		string fob = "";
		string fora = "";
		for (int j = 0; j < n; j++) {
			forb += (char)(b[i][j] + '0');
			fob += (char)(!b[i][j] + '0');
			fora += (char)(a[i][j] + '0');
		}
		if (fora != forb && fora != fob) {
			flag = 0; break;
		}


	}
	if (flag) {
		cout << "YES" << endl; return;
	}
	// flip first row
	for (auto x : col) {
		for (int i = 0; i < n; i++) {
			a[i][x] = !a[i][x];
		}
	}
	vi co;
	for (int i = 0; i < n; i++) {
		a[0][i] = !a[0][i];
		if (a[0][i] != b[0][i])co.pb(i);
	}
	for (auto x : col) {
		for (int i = 0; i < n; i++) {
			a[i][x] = !a[i][x];
		}
	}
	for (int i = 0; i < n; i++) {
		string forb = "", forbb = "";
		string fora = "";
		for (int j = 0; j < n; j++) {
			forb += (char)(b[i][j] + '0');
			forbb += (char)(!b[i][j] + '0');
			fora += (char)(a[i][j] + '0');
		}
		if (fora != forb && fora != forbb) {
			flag = 0; break;
		}


	}
	if (flag) {
		cout << "YES" << endl; return;
	}
	else {
		cout << "NO" << endl;
	}


}

bool TestCase = 1;


//////////////////////////////////////////////////////////////////




int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	ll t;
	t = 1;
	if (TestCase) {
		cin >> t;
	}
	while (t--) {
		if (isGoogles) {cout << "Case #" << caseNumber << ": ";} caseNumber++;
		jabru();
	}
	return 0;
}