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

public class Main {

    void run() throws IOException {
        int n = nextInt();
        char[] s1 = nextToken().toCharArray();
        char[] s2 = nextToken().toCharArray();
        int c1 = 0;
        int c2 = 0;
        for (int i = 0; i < n; i++) {
            c1 += s1[i] - '0';
            c2 += s2[i] - '0';
        }
        if (c1 != c2) out.println("no");
        else {
            boolean wast = false;
            boolean wasf = false;
            int count = 0;
            for (int i = 0; i < n; i++) {
                wast |= s1[i] == s2[i];
                wasf |= s1[i] != s2[i];
                if (s1[i] == '1') count++;
                else count--;
                if (count == 0 && wasf && wast) {
                    out.println("no");
                    return;
                }
                if (count == 0) wasf = wast = false;
            }
            if(wasf && (count != 0 || wast)) out.println("no");
            else out.println("yes");
        }
    }

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(System.out);
    StringTokenizer in = new StringTokenizer("");


    // pairs
    class pil implements Comparable<pil> {
        int first;
        long second;

        pil(int f, long s) {
            this.first = f;
            this.second = s;
        }

        public int compareTo(pil o) {
            if (first != o.first) return Integer.compare(first, o.first);
            return Long.compare(second, o.second);
        }
    }

    class pii implements Comparable<pii> {
        int se;
        int fi;

        pii(int s, int f) {
            this.se = f;
            this.fi = s;
        }

        public int compareTo(pii o) {
            if (fi != o.fi) return Integer.compare(fi, o.fi);
            return Integer.compare(se, o.se);
        }
    }

    class pll implements Comparable<pll> {
        long first;
        long second;

        pll(long f, long s) {
            this.first = f;
            this.second = s;
        }

        public int compareTo(pll o) {
            if (first != o.first) return Long.compare(first, o.first);
            return Long.compare(second, o.second);
        }
    }

    class pli implements Comparable<pli> {
        long first;
        int second;

        pli(long f, int s) {
            this.first = f;
            this.second = s;
        }

        public int compareTo(pli o) {
            if (first != o.first) return Long.compare(first, o.first);
            return Integer.compare(second, o.second);
        }
    }

    boolean hasNext() throws IOException {
        if (in.hasMoreTokens()) return true;
        String s;
        while ((s = br.readLine()) != null) {
            in = new StringTokenizer(s);
            if (in.hasMoreTokens()) return true;
        }
        return false;
    }

    String nextToken() throws IOException {
        while (!in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public static void main(String[] args) throws IOException {
        Main m = new Main();
        int t = m.nextInt();
        while (t-- > 0)
            m.run();
        m.out.close();
    }
}