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.
//Author: AnandRaj doubleux
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef pair<ll,ll> pll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<pll> vpll; 

#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define test() int t;cin>>t;while(t--)
#define all(v) v.begin(),v.end()
#define prinp(p) cout<<p.first<<" "<<p.second<<endl
#define prinv(V) for(auto v:V) cout<<v<<" ";cout<<endl
#define take(V,f,n) for(int in=f;in<f+n;in++) cin>>V[in]
#define what(x) cerr<<#x<<" = "<<x<<endl
#define KStest() int t,t1;cin>>t;t1=t;while(t--)
#define KScout cout<<"Case #"<<t1-t<<": "
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1e9+7,MAX = 1e6+5;
const ll INF = 1e18+5;

bool fun(vector<string> &A, vector<string> &B)
{
	int n = A.size();
	if(n==0) return true;
	bool can = false;

	vector<string> XB(n-1),XA(n-1);
	for(int i=1;i<n;i++)
	{
		for(int j=1;j<n;j++)
		{
			XB[i-1].push_back(B[i][j]);
			XA[i-1].push_back(A[i][j]);
		}
	}

	if(A[0][0]==B[0][0])
	{
		
		vector<string> XXB = XB;
		vector<string> XXA = XA;
		for(int i=1;i<n;i++)
		{
			if(A[0][i]!=B[0][i])
			{
				for(int j=1;j<n;j++)
				{
					XXA[j-1][i-1] = '0' + (1-(XXA[j-1][i-1]-'0'));
				}
			}
			if(A[i][0]!=B[i][0])
			{
				for(int j=1;j<n;j++)
				{
					XXA[i-1][j-1] = '0' + (1-(XXA[i-1][j-1]-'0'));
				}
			}
		}
		can = (XXA == XXB);
		// cerr<<can<<" ";
		XXA = XA;
		XXB = XB;
		for(int i=1;i<n;i++)
		{
			if(A[0][i]==B[0][i])
			{
				for(int j=1;j<n;j++)
				{
					XXA[j-1][i-1] = '0' + (1-(XXA[j-1][i-1]-'0'));
				}
			}
			if(A[i][0]==B[i][0])
			{
				for(int j=1;j<n;j++)
				{
					XXA[i-1][j-1] = '0' + (1-(XXA[i-1][j-1]-'0'));
				}
			}
		}
		can = (XXA == XXB);
		// cerr<<can<<" ";
	}
	else
	{
		vector<string> XXB = XB;
		vector<string> XXA = XA;
		for(int i=1;i<n;i++)
		{
			if(A[0][i]==B[0][i])
			{
				for(int j=1;j<n;j++)
				{
					XXA[j-1][i-1] = '0' + (1-(XXA[j-1][i-1]-'0'));
				}
			}
			if(A[i][0]!=B[i][0])
			{
				for(int j=1;j<n;j++)
				{
					XXA[i-1][j-1] = '0' + (1-(XXA[i-1][j-1]-'0'));
				}
			}
		}
		can = can = (XXA == XXB);
		// cerr<<can<<" ";
		XXA = XA;
		XXB = XB;
		for(int i=1;i<n;i++)
		{
			if(A[0][i]!=B[0][i])
			{
				for(int j=1;j<n;j++)
				{
					XXA[j-1][i-1] = '0' + (1-(XXA[j-1][i-1]-'0'));
				}
			}
			if(A[i][0]==B[i][0])
			{
				for(int j=1;j<n;j++)
				{
					XXA[i-1][j-1] = '0' + (1-(XXA[i-1][j-1]-'0'));
				}
			}
		}

		can = can = (XXA == XXB);
		// cerr<<can<<" ";
	}
	// cout<<endl;
	return can;
}

int main()
{
	test()
	{
		int n;
		cin>>n;
		vector<string> A(n),B(n);
		take(A,0,n);
		take(B,0,n);
		bool can = fun(A,B);
		if(can) cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
}	