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 Solution {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader(String s) {
            try {
                br = new BufferedReader(new FileReader(s));
            } catch (FileNotFoundException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }

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

        String nextToken() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

        String nextLine() throws IOException {
            return br.readLine();
        }

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

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

        float nextFloat() {
            return Float.parseFloat(nextToken());
        }
    }


    static FastReader f = new FastReader();
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st;
    static StringBuilder sb = new StringBuilder("");
    private static int mod = 998244353;
    static int MAX = 500005;
    static long[] fact;

    static int[] inputArray(int n) throws IOException {
        int[] a = new int[n];
        for(int i = 0 ; i < n ; i++) {
//            a[i] = (int) (Math.random()*1e5 + 1);
            a[i] = f.nextInt();
        }
        return a;
    }
    static long[] inputLongArray(int n) throws IOException {
        long[] a = new long[n];
        for(int i = 0 ; i < n ; i++) {
            a[i] = f.nextLong();
//            a[i] = (long) (Math.random() * 1e9 + 1);
        }
        return a;
    }

    static int gcd(int a , int b) {
        if(a == 0 || b == 0) {
            return Math.max(a , b);
        }
        //System.out.println("a - " + a + " b - " + b);
        if(a % b == 0) {
            return b;
        }
        return gcd(b , a % b);
    }

    static void initializeFact() {
        fact = new long[MAX];
        for(int i = 0 ; i < fact.length ; i++) {
            if(i == 0) {
                fact[i] = 1;
            }
            else {
                fact[i] = fact[i-1] * i % mod;
            }
        }
    }

    static long longModulus(long x , long m) {
        if(x < m) {
            return x;
        }
        long d = x / m;
        return x - d * m;
    }


    static boolean[] sieveOfEratosthenes(int n)
    {
        boolean[] isPrime = new boolean[n+1];
        Arrays.fill(isPrime , true);
        isPrime[0] = false;
        isPrime[1] = false;
        for(int i = 2; i * i <= n ; i++)
        {
            for(int j = 2 * i ; j <= n; j += i)
                isPrime[j] = false;
        }
        return isPrime;
    }

    static long moduloInversePrime(long a) {
        //System.out.println("modulo inverse of " + a + " -> " + ans);
        return modPow(a , mod - 2);
    }

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

    static long modPow(long a, int step)
    {
        long ans = 1;
        while(step != 0)
        {
            if((step & 1) != 0)
                ans = mult(ans , a);
            a = mult(a , a);
            step >>= 1;
        }
        return ans;
    }


    public static void main(String[] args) throws IOException {
        int test = f.nextInt();
        for(int t = 1 ; t <= test ; t++) {
            int n = f.nextInt();
            String a = f.nextToken();
            String b = f.nextToken();
            int[] bits = new int[n];
            for(int i = 0 ; i < n ; i++) {
                if(b.charAt(i) != a.charAt(i)) {
                    bits[i] = 1;
                }
            }
            int[] posible = new int[n];
            int ones = 0;
            for(int i = 0 ; i < n ; i++) {
                if(a.charAt(i) == '1') {
                    ones++;
                }
                if((i + 1) % 2 == 0 && ones  == (i + 1) / 2) {
                    posible[i] = 1;
                }
            }

            boolean is_possible = true;
            int prev = -1;
            for(int i = n-1 ; i >= 0 ; i--) {
                if(posible[i] == 1) {
                    break;
                }
                if(bits[i] == 1) {
                    is_possible = false;
                    break;
                }
            }
            if(!is_possible) {
//                System.out.println("1");
                System.out.println("NO");
                continue;
            }
//            System.out.println(Arrays.toString(bits));
//            System.out.println(Arrays.toString(posible));
            for(int i = 1 ; i < n ; i++) {
                if(posible[i-1] == 0 && bits[i] != bits[i-1]) {
//                    System.out.println("i - " + i);
                    is_possible = false;
                    break;
                }
            }
            if(!is_possible) {
//                System.out.println("2");
                System.out.println("NO");
            }
            else {
                System.out.println("YES");
            }
        }
    }


}

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