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.*;
import java.math.*;

public class CFTemplate {
    static final long MOD = 1000000007L;
    //static final long MOD2 = 1000000009L;
    //static final long MOD = 998244353L;
    //static final long INF = 500000000000L;
    static final int INF = 10000000;
    static final int NINF = -100000;
    static FastScanner sc;
    static PrintWriter pw;

    static int N;
    static int[][] A;
    static int[][] B;
    public static void main(String[] args) {
        sc = new FastScanner();
        pw = new PrintWriter(System.out);

        int Q = sc.ni();
        for (int q = 0; q < Q; q++) {
            N = sc.ni();
            A = new int[N][N];
            for (int i = 0; i < N; i++) {
                String row = sc.next();
                for (int j = 0; j < N; j++)
                    A[i][j] = row.charAt(j)-'0';
            }
            B = new int[N][N];
            for (int i = 0; i < N; i++) {
                String row = sc.next();
                for (int j = 0; j < N; j++)
                    B[i][j] = row.charAt(j)-'0';
            }
            
            boolean ans = solve(true)|solve(false);
            if (ans)
                pw.println("YES");
            else
                pw.println("NO");
            
        }
        pw.close();
    }

    public static boolean solve(boolean flip) {
        int[][] mat = A.clone();
        if (flip) {
            for (int i = 0; i < N; i++)
                mat[i][0] = 1-mat[i][0];
        }
        for (int i = 0; i < N; i++) {
            if (mat[i][0] != B[i][0]) {
                for (int j = 0; j < N; j++)
                    mat[i][j] = 1-mat[i][j];
            }
        }
        for (int j = 1; j < N; j++) {
            if (mat[0][j] != B[0][j]) {
                for (int i = 0; i < N; i++)
                    mat[i][j] = 1-mat[i][j];
            }
        }

        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                if (mat[i][j] != B[i][j])
                    return false;
            }
        }
        return true;
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in), 32768);
            st = null;
        }
 
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int ni() {
            return Integer.parseInt(next());
        }

        int[] intArray(int N) {
            int[] ret = new int[N];
            for (int i = 0; i < N; i++)
                ret[i] = ni();
            return ret;
        }
 
        long nl() {
            return Long.parseLong(next());
        }

        long[] longArray(int N) {
            long[] ret = new long[N];
            for (int i = 0; i < N; i++)
                ret[i] = nl();
            return ret;
        }

        double nd() {
            return Double.parseDouble(next());
        }
 
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}