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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author sarthakmanna
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        FUnusualMatrix solver = new FUnusualMatrix();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class FUnusualMatrix {
        public void solve(int testNumber, FastReader in, PrintWriter out) {
            int i, j, k;

            int N = in.nextInt();
            String[] A = in.getStringArray(N);
            String[] B = in.getStringArray(N);

            int[][] isSame = new int[N][N];
            for (i = 0; i < N; ++i)
                for (j = 0; j < N; ++j) {
                    isSame[i][j] = A[i].charAt(j) == B[i].charAt(j) ? 0 : 1;
                }

            for (i = 0; i < 2; ++i)
                for (j = 0; j < 2; ++j) {
                    if ((isSame[0][0] ^ i ^ j) == 0) {
                        int[] row = new int[N], col = new int[N];

                        row[0] = i;
                        col[0] = j;
                        for (k = 1; k < N; ++k) {
                            row[k] = col[0] ^ isSame[k][0];
                            col[k] = row[0] ^ isSame[0][k];
                        }

                        boolean ok = true;
                        for (int x = 0; x < N; ++x)
                            for (int y = 0; y < N; ++y) {
                                if ((row[x] ^ col[y] ^ isSame[x][y]) != 0) ok = false;
                            }
                        if (ok) {
                            out.println("YES");
                            return;
                        }
                    }
                }
            out.println("NO");
        }

    }

    static class FastReader {
        static final int BUFSIZE = 1 << 20;
        static byte[] buf;
        static int index;
        static int total;
        static InputStream in;

        public FastReader(InputStream is) {
            try {
                in = is;
                buf = new byte[BUFSIZE];
            } catch (Exception e) {
            }
        }

        private int scan() {
            try {
                if (index >= total) {
                    index = 0;
                    total = in.read(buf);
                    if (total <= 0)
                        return -1;
                }
                return buf[index++];
            } catch (Exception | Error e) {
                System.err.println(e.getMessage());
                return 13 / 0;
            }
        }

        public String next() {
            int c;
            for (c = scan(); c <= 32; c = scan()) ;
            StringBuilder sb = new StringBuilder();
            for (; c > 32; c = scan())
                sb.append((char) c);
            return sb.toString();
        }

        public int nextInt() {
            int c, val = 0;
            for (c = scan(); c <= 32; c = scan()) ;
            boolean neg = c == '-';
            if (c == '-' || c == '+')
                c = scan();
            for (; c >= '0' && c <= '9'; c = scan())
                val = (val << 3) + (val << 1) + (c & 15);
            return neg ? -val : val;
        }

        public String[] getStringArray(int size) {
            String[] ar = new String[size];
            for (int i = 0; i < size; ++i) ar[i] = next();
            return ar;
        }

    }
}

