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 F {

	static long m = (long) (1e9 + 7);

	public static void main(String[] args) throws IOException {
		Scanner scn = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		StringBuilder sb = new StringBuilder();
		int T = scn.ni(), tcs = 0;
		F: while (tcs++ < T) {
			int n = scn.ni();
			int m[][] = new int[n][n];
			for (int i = 0; i < n; i++) {
				char s[] = scn.next().toCharArray();
				for (int j = 0; j < n; j++)
					m[i][j] = (s[j] - '0');
			}

			int m2[][] = new int[n][n];
			for (int i = 0; i < n; i++) {
				char s[] = scn.next().toCharArray();
				for (int j = 0; j < n; j++)
					m2[i][j] = (s[j] - '0');
			}
			for (int j = 0; j < n; j++) {
				if (m[0][j] != m2[0][j]) {
					for (int i = 0; i < n; i++) {
						m[i][j] ^= 1;
					}
				}
			}
			boolean b = true;
			for (int i = 1; i < n; i++) {
				int cnt = 0;
				for (int j = 0; j < n; j++) {
					if (m[i][j] != m2[i][j])
						cnt++;
				}
				if (cnt > 0 && cnt < n) {
					b = false;
					break;
				}
			}
			sb.append(b ? "YES" : "NO");
			sb.append("\n");
		}
		out.print(sb);
		out.close();
	}

	static class Scanner {

		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int ni() throws IOException {
			return Integer.parseInt(next());
		}

		public long nl() throws IOException {
			return Long.parseLong(next());
		}

		public int[] nia(int n) throws IOException {
			int a[] = new int[n];
			String sa[] = br.readLine().split(" ");
			for (int i = 0; i < n; i++)
				a[i] = Integer.parseInt(sa[i]);
			return a;
		}

		public long[] nla(int n) throws IOException {
			long a[] = new long[n];
			String sa[] = br.readLine().split(" ");
			for (int i = 0; i < n; i++)
				a[i] = Long.parseLong(sa[i]);
			return a;
		}

		public int[] sort(int[] a) {
			ArrayList<Integer> l = new ArrayList<>();
			for (int v : a)
				l.add(v);
			Collections.sort(l);
			for (int i = 0; i < a.length; i++)
				a[i] = l.get(i);
			return a;
		}

		public long[] sort(long[] a) {
			ArrayList<Long> l = new ArrayList<>();
			for (long v : a)
				l.add(v);
			Collections.sort(l);
			for (int i = 0; i < a.length; i++)
				a[i] = l.get(i);
			return a;
		}
	}
}