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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class B {

    static final long mod = (long) 1e9 + 7l;
    static int z = 0, o = 0;

    private static void solve(int t) {

        int n = fs.nextInt();
        char inp[] = fs.next().toCharArray();
        char ot[] = fs.next().toCharArray();
        int cnt = 0, last = 0;
        for (int i = 0; i < n; i++) {
            if (inp[i] == '1') cnt++;
            else cnt--;
            if (cnt == 0) {
                int f1 = 0, f2 = 0;
                for (int j = last; j <= i; j++) {
                    if (inp[j] != ot[j]) f1 = 1;
                    else f2 = 1;
                }
                if (f1 == 1 && f2 == 1) {
                    out.println("NO");
                    return;
                }
                last = i + 1;
            }
        }
        if (last != n) {
            for (int i = last; i < n; i++) {
                if (inp[i] != ot[i]) {
                    out.println("No");
                    return;
                }
            }
        }
        out.println("Yes");
    }

    private static void flip(char[] inp, int idx) {
        if (inp[idx] == '1') {
            o++;
            inp[idx] = '1';
        } else {
            z++;
            inp[idx] = '0';
        }
    }


    public static void main(String[] args) {
        fs = new FastScanner();
        out = new PrintWriter(System.out);
        long s = System.currentTimeMillis();
        int t = fs.nextInt();
        for (int i = 1; i <= t; i++) solve(t);
        out.close();
//        System.err.println( System.currentTimeMillis() - s + "ms" );
    }

    static boolean DEBUG = true;
    static PrintWriter out;
    static FastScanner fs;

    static void trace(Object... o) {
        if (!DEBUG) return;
        System.err.println(Arrays.deepToString(o));
    }

    static void pl(Object o) {
        out.println(o);
    }

    static void p(Object o) {
        out.print(o);
    }

    static long gcd(long a, long b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static int gcd(int a, int b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static void sieveOfEratosthenes(int n, int factors[]) {
        factors[1] = 1;
        for (int p = 2; p * p <= n; p++) {
            if (factors[p] == 0) {
                factors[p] = p;
                for (int i = p * p; i <= n; i += p)
                    factors[i] = p;
            }
        }
    }

    static long mul(long a, long b) {
        return a * b % mod;
    }

    static long fact(int x) {
        long ans = 1;
        for (int i = 2; i <= x; i++) ans = mul(ans, i);
        return ans;
    }

    static long fastPow(long base, long exp) {
        if (exp == 0) return 1;
        long half = fastPow(base, exp / 2);
        if (exp % 2 == 0) return mul(half, half);
        return mul(half, mul(half, base));
    }

    static long modInv(long x) {
        return fastPow(x, mod - 2);
    }

    static long nCk(int n, int k) {
        return mul(fact(n), mul(modInv(fact(k)), modInv(fact(n - k))));
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

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