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.
import java.io.*;
import java.util.*;

public class CodeForces {

	public static void main(String[] args) throws FileNotFoundException {
		FastScanner sc = new FastScanner();
		StringBuilder sb = new StringBuilder();
		int t = sc.nextInt();
		while (t-- > 0) {
			solve(sc, sb);
		}
		System.out.print(sb);
	}

	

	private static void solve(FastScanner sc, StringBuilder sb) {
		int n=sc.nextInt();
		int[][] a=new int[n][n];
		int[][] b=new int[n][n];
		for(int i=0;i<n;i++) {
			String str=sc.next();
			for (int j = 0; j < n; j++) {
				a[i][j]=str.charAt(j)-'0';
			}
		}
		
		for(int i=0;i<n;i++) {
			String str=sc.next();
			for (int j = 0; j < n; j++) {
				b[i][j]=str.charAt(j)-'0';
			}
		}
		
		for(int i=0;i<n;i++) {
			if(a[0][i]!=b[0][i]) {
				for(int j=0;j<n;j++) {
					a[j][i]^=1;
				}
			}
		}
		for(int i=0;i<n;i++) {
			if(a[i][0]!=b[i][0]) {
				for(int j=0;j<n;j++) {
					a[i][j]^=1;
				}
			}
		}
		for (int i = 1; i < b.length; i++) {
			for (int j = 1; j < b.length; j++) {
				if(a[i][j]!=b[i][j]) {
					sb.append("NO\n");return;
				}
			}
		}
		sb.append("YES\n");

	}

}

class Pair implements Comparable<Pair> {
	int r, c;

	public Pair(int i, int j) {
		r = i;
		c = j;// TODO Auto-generated constructor stub

	}

	@Override
	public int compareTo(Pair o) {

		return o.c == c ? r - o.r : c - o.c;
	}
}

class FastScanner {
	BufferedReader br;
	StringTokenizer st;

	public FastScanner() throws FileNotFoundException {
		try {
			File file = new File("src\\input.txt");
			br = new BufferedReader(new FileReader(file));
		} catch (Exception e) {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

	}

	String next() {
		while (st == null || !st.hasMoreElements()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

	int nextInt() {
		return Integer.parseInt(next());
	}

	long nextLong() {
		return Long.parseLong(next());
	}

	double nextDouble() {
		return Double.parseDouble(next());
	}

	int[] nextInts(int N) {
		int[] res = new int[N];
		for (int i = 0; i < N; i++) {
			res[i] = (int) nextLong();
		}
		return res;
	}

	long[] nextLongs(int N) {
		long[] res = new long[N];
		for (int i = 0; i < N; i++) {
			res[i] = nextLong();
		}
		return res;
	}

	public double[] nextDoubles(int N) {
		double[] res = new double[N];
		for (int i = 0; i < N; i++) {
			res[i] = nextDouble();
		}
		return res;
	}

	String nextLine() {
		String str = "";
		try {
			str = br.readLine();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return str;
	}
}
