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.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
//https://codeforces.com/contest/1475/problem/F

public class UnusualMatrix {

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int T = Integer.parseInt(br.readLine());
		
		while(T-- > 0) {
			if(solve(br)) {
				System.out.println("Yes");
			}else {
				System.out.println("No");
			}	
		}
	}
	
	private static boolean solve(BufferedReader br) throws IOException {
		int N = Integer.parseInt(br.readLine());
		int A[][] = new int[N][N];
		int B[][] = new int[N][N];
		for(int i = 0 ; i < N ; i++) {
			String s = br.readLine();
			for(int j = 0 ; j < N ; j++) {
				A[i][j] = (int) s.charAt(j) - 48;
			}
		}
		br.readLine();
		for(int i = 0 ; i < N ; i++) {
			String s = br.readLine();
			for(int j = 0 ; j < N ; j++) {
				B[i][j] = (int) s.charAt(j) - 48;
			}
		}
		boolean allXORed = true;
		boolean allSame = true;
		
		for(int i = 0 ; i < N ; i++) {
			if(A[0][i] != B[0][i]) {
				allSame = false;
			} else {
				allXORed = false;
			}
		}
		
		if(!allSame && !allXORed) {
			for(int i = 0 ; i < N ; i++) {
				if(A[0][i] != B[0][i]) {
					for(int j = 0 ; j < N ; j++) {
						A[j][i] = A[j][i] ^ 1;
					}
				}
			}
		}
		
		for(int i = 1 ; i < N ; i++) {
			boolean allXor = true;
			boolean allEq = true;
			for(int j = 0 ; j < N ; j++) {
				if(A[i][j] != (B[i][j] ^ 1)) {
					allXor = false;
				}else if(A[i][j] != B[i][j]) {
					allEq = false;
				}
			}
			if(!allXor && !allEq) return false;
		}
		
		return true;
	}
	 static class Reader {
	        final private int BUFFER_SIZE = 1 << 16;
	        private DataInputStream din;
	        private byte[] buffer;
	        private int bufferPointer, bytesRead;
	 
	        public Reader()
	        {
	            din = new DataInputStream(System.in);
	            buffer = new byte[BUFFER_SIZE];
	            bufferPointer = bytesRead = 0;
	        }
	 
	        public Reader(String file_name) throws IOException
	        {
	            din = new DataInputStream(
	                new FileInputStream(file_name));
	            buffer = new byte[BUFFER_SIZE];
	            bufferPointer = bytesRead = 0;
	        }
	 
	        public String readLine() throws IOException
	        {
	            byte[] buf = new byte[64]; // line length
	            int cnt = 0, c;
	            while ((c = read()) != -1) {
	                if (c == '\n') {
	                    if (cnt != 0) {
	                        break;
	                    }
	                    else {
	                        continue;
	                    }
	                }
	                buf[cnt++] = (byte)c;
	            }
	            return new String(buf, 0, cnt);
	        }
	 
	        public int nextInt() throws IOException
	        {
	            int ret = 0;
	            byte c = read();
	            while (c <= ' ') {
	                c = read();
	            }
	            boolean neg = (c == '-');
	            if (neg)
	                c = read();
	            do {
	                ret = ret * 10 + c - '0';
	            } while ((c = read()) >= '0' && c <= '9');
	 
	            if (neg)
	                return -ret;
	            return ret;
	        }
	 
	        public long nextLong() throws IOException
	        {
	            long ret = 0;
	            byte c = read();
	            while (c <= ' ')
	                c = read();
	            boolean neg = (c == '-');
	            if (neg)
	                c = read();
	            do {
	                ret = ret * 10 + c - '0';
	            } while ((c = read()) >= '0' && c <= '9');
	            if (neg)
	                return -ret;
	            return ret;
	        }
	 
	        public double nextDouble() throws IOException
	        {
	            double ret = 0, div = 1;
	            byte c = read();
	            while (c <= ' ')
	                c = read();
	            boolean neg = (c == '-');
	            if (neg)
	                c = read();
	 
	            do {
	                ret = ret * 10 + c - '0';
	            } while ((c = read()) >= '0' && c <= '9');
	 
	            if (c == '.') {
	                while ((c = read()) >= '0' && c <= '9') {
	                    ret += (c - '0') / (div *= 10);
	                }
	            }
	 
	            if (neg)
	                return -ret;
	            return ret;
	        }
	 
	        private void fillBuffer() throws IOException
	        {
	            bytesRead = din.read(buffer, bufferPointer = 0,
	                                 BUFFER_SIZE);
	            if (bytesRead == -1)
	                buffer[0] = -1;
	        }
	 
	        private byte read() throws IOException
	        {
	            if (bufferPointer == bytesRead)
	                fillBuffer();
	            return buffer[bufferPointer++];
	        }
	 
	        public void close() throws IOException
	        {
	            if (din == null)
	                return;
	            din.close();
	        }
	    }
}
