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.util.*;
import java.io.*;

public class _696 {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int [][] a = new int[n][n];
            for (int i = 0; i < n; i++) {
                String s = sc.next();
                for (int j = 0; j < n; j++) {
                    a[i][j] = s.charAt(j) - '0';
                }
            }
            int [][] b = new int[n][n];
            for (int i = 0; i < n; i++) {
                String s = sc.next();
                for (int j = 0; j < n; j++) {
                    b[i][j] = s.charAt(j) - '0';
                    a[i][j] = a[i][j] ^ b[i][j];
                }
            }
            boolean res = false;
            for (int k = 0; k < 2; k++) {
                int [] row = new int[n];
                int [] col = new int[n];
                Arrays.fill(row, -1); Arrays.fill(col, -1);
                row[0] = k;
                for (int j = 0; j < n; j++) {
                    col[j] = row[0] ^ a[0][j];
                }
                for (int i = 1; i < n; i++) {
                    row[i] = col[0] ^ a[i][0];
                }
                boolean ok = true;
                for (int i = 0; i < n; i++) {
                    for (int j = 0; j < n; j++) ok = ok && ((row[i] ^ col[j]) == a[i][j]);
                }
                res = res || ok;
            }
            out.println(res ? "YES" : "NO");
        }
        out.close();
    }

    static long[] fac;
    static long mod = (long) 1e9 + 7;

    static void initFac(long n) {
        fac = new long[(int)n + 1];
        fac[0] = 1;
        for (int i = 1; i <= n; i++) {
            fac[i] = (fac[i - 1]  * i) % mod;
        }
    }



    static long nck(int n, int k) {
        if (n < k)
            return 0;
        long den = inv((int) (fac[k] * fac[n - k] % mod));
        return fac[n] * den % mod;
    }

    static long pow(long b, long e) {
        long ans = 1;
        while (e > 0) {
            if (e % 2 == 1)
                ans = ans * b % mod;
            e >>= 1;
            b = b * b % mod;
        }
        return ans;
    }

    static long inv(long x) {
        return pow(x, mod - 2);
    }




    static void sort(int[] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i : a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }

    static void sort(long[] a) {
        ArrayList<Long> q = new ArrayList<>();
        for (long i : a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }


    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }


    }

}