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.
// Code with Pikachu

#include <iostream>
#include <vector>
#include <algorithm>
#include <chrono>
#include <random>
#include <map>
#include <unordered_map>
#include <stack>
#include <queue>
#include <math.h>
#include <ctype.h>
#include <cstring>
#include <string>
//#include <bits/stdc++.h>
using namespace std;

#define fo(i, n) for (i = 0; i < n; i++)
#define Fo(i,s,n) for(i = s; i < n; i++)
#define loop(i,s,n,itr) for(i = s; i < n; i += itr)
#define ll long long
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define MI map<int,int>
#define MC map<char,int>
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for (auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
#define endl "\n"
#define deb(x) cout << #x << "=" << x << "\n"
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << "\n"
#define deb3(x, y, z) cout << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << "\n"
#define deb4(x, y, z, w) cout << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << "," << #w << "=" << w << "\n"
#define input(arr,n) for(int i=0;i<n;i++) cin>>arr[i]
#define showarr(arr,n) for(int  i=0; i<n; i++) cout<<arr[i]<<" "; cout<<endl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef vector<pl> vpl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());

const int mod = 1000000007;
const int N = 3e5, M = N;
//=======================o

vi g[N];

void solve()
{
	int i = 0, j = 0, n, m, k, flag = 0 , ans = 0;
	cin >> n;
	int a[n][n], b[n][n], cng[n][n], row[n], col[n];
	clr(row);
	clr(col);
	string s;
	fo(i, n)
	{
		cin >> s;
		fo(j, n)
		a[i][j] = s[j] - '0';
	}
	fo(i, n)
	{
		cin >> s;
		fo(j, n)
		b[i][j] = s[j] - '0';
	}
	fo(j, n)
	{
		if (a[0][j] != b[0][j])
		{
			fo(i, n)
			a[i][j] ^= 1;
		}
	}
	Fo(i, 1, n)
	{
		flag = 0;
		fo(j, n)
		{
			if (a[i][j] != b[i][j])
			{
				flag = 1;
				break;
			}
		}
		if (flag)
		{
			fo(j, n)
			{
				a[i][j] ^= 1;
			}
		}
		fo(j, n)
		{
			if (a[i][j] != b[i][j])
			{
				cout << "NO" << endl;
				return;
			}
		}

	}
	cout << "YES" << endl;
}
int main()
{
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	srand(chrono::high_resolution_clock::now().time_since_epoch().count());
	int t = 1;
	// freopen("input.txt", "r", stdin);
	// freopen("output.txt", "w", stdout);
	cin >> t;
	while (t--)
	{
		solve();
	}
	return 0;
}