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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.ArrayList;

public class F {
	public void solve(IO io) throws IOException {
		int n = io.nextInt();
		int[][] a = new int[n][n];
		Raw[] rows = new Raw[n];
		Raw[] cols = new Raw[n];
		for (int i = 0; i < n; ++i) {
			String s = io.nextToken();
			for (int j = 0; j < n; ++j) {
				if (i == 0) {
					a[j] = new int[n];
				}
				a[i][j] = (int)s.charAt(j) - 48;
			}
		}
		for (int i = 0; i < n; ++i) {
			int[] row = new int[n];
			String s = io.nextToken();
			for (int j = 0; j < n; ++j) {
				a[i][j] ^= ((int)s.charAt(j) - 48);
				row[j] = a[i][j];
			}
			rows[i] = new Raw(row);
		}
		for (int j = 0; j < n; ++j) {
			int[] col = new int[n];
			for (int i = 0; i < n; ++i) {
				col[i] = a[i][j];
			}
			cols[j] = new Raw(col);
		}
		boolean fr = true;
		boolean fc = true;
		for (int i = 1; i < n; ++i) {
			fr &= rows[i - 1].equals(rows[i]);
			fc &= cols[i - 1].equals(cols[i]);
		}
		if (fr || fc) {
			io.println("YES");
		} else {
			io.println("NO");
		}
	}

    public void run() throws IOException {
		IO io = new IO();
        int t = io.nextInt();
		for (int i = 0; i < t; ++i) {
			solve(io);
		}
		io.close();
    }

    public static void main(String[] args) throws IOException {
        new F().run();
    }

}

class IO {
	public IO() {
        br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
    }

    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;

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

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

	public void print(String a) {
		out.print(a);
	}

	public void println(String a) {
		out.println(a);
	}

	public void close() {
		out.close();
	}
}

class Raw {
	int[] a;
	Raw() {}
	Raw(int[] a1) {
		this.a = a1;
	}
	boolean equals(Raw b) {
		boolean eq = true;
		boolean ueq = true;
		for (int i = 0; i < a.length; ++i) {
			eq &= (a[i] == b.a[i]);
			ueq &= (a[i] != b.a[i]);
		}
		return eq | ueq;
	}
}
