There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.

For example, suppose a=0111010000. 

  * In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00. 
  * In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100. 
  * It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1. 



Can you transform the string a into the string b using some finite number of operations (possibly, none)?

Input

The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.

The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.

The sum of n across all test cases does not exceed 3⋅ 10^5.

Output

For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).

Example

Input


5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100


Output


YES
YES
NO
YES
NO

Note

The first test case is shown in the statement.

In the second test case, we transform a into b by using zero operations.

In the third test case, there is no legal operation, so it is impossible to transform a into b.

In the fourth test case, here is one such transformation: 

  * Select the length 2 prefix to get 100101010101. 
  * Select the length 12 prefix to get 011010101010. 
  * Select the length 8 prefix to get 100101011010. 
  * Select the length 4 prefix to get 011001011010. 
  * Select the length 6 prefix to get 100110011010. 



In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
import java.util.*;
import java.io.*;
import java.lang.*;

public class Solution {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(System.out);
    StringTokenizer st = new StringTokenizer("");

    String next() {
        if (!st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {

            }
        }
        return st.nextToken();
    }

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

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

    public static void main(String[] args) {
        new Solution().solve();
    }

    void solve() {
        int t = 1;
        t = nextInt();
        for (int tt = 0; tt < t; tt++) {
            int n = nextInt();
            char arr[] = next().toCharArray();
            char brr[] = next().toCharArray();
            Point p[] = new Point[n];
            p[0] = new Point(0, 0);

            if (arr[0] == '0') {
                p[0].z++;
            } else {
                p[0].o++;
            }

            for (int i = 1; i < n; i++) {
                p[i] = new Point(p[i - 1].z, p[i - 1].o);
                if (arr[i] == '0') {
                    p[i].z++;
                } else {
                    p[i].o++;
                }
            }
            boolean flag = true, changed = false;
            for (int i = n - 1; i >= 0; i--) {
                if (!changed) {
                    if (arr[i] != brr[i]) {
                        if (p[i].z != p[i].o) {
                            flag = false;
                        } else {
                            changed = true;
                        }
                    }
                } else {
                    if (arr[i] == brr[i]) {
                        if (p[i].z != p[i].o) {
                            flag = false;
                        } else {
                            changed = false;
                        }
                    }
                }
            }
            // for (Point a : p) {
            // out.println(a);
            // }
            out.println(flag ? "YES" : "NO");
        }
        out.close();
    }
}

class Point {
    int z = 0, o = 0;

    Point(int z, int o) {
        this.z = z;
        this.o = o;
    }

    @Override
    public String toString() {
        return this.z + " " + this.o;
    }
}