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.*;
import java.lang.*;
import java.text.*;

public class B {

    static int M = 1000000007;

    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    public static int nextInt() throws NumberFormatException, IOException {
        return Integer.parseInt(br.readLine());
    }

    public static long nextLong() throws NumberFormatException, IOException {
        return Long.parseLong(br.readLine());
    }

    public static String next() throws IOException {
        return br.readLine();
    }

    public static int[] getArray(int n) throws NumberFormatException, IOException {
        int[] a = new int[n];
        String[] str = br.readLine().split(" ");
        for (int i = 0; i < n; i++)
            a[i] = Integer.parseInt(str[i]);
        return a;
    }

    public static long[] getLongArray(int n) throws NumberFormatException, IOException {
        long[] a = new long[n];
        String[] str = br.readLine().split(" ");
        for (int i = 0; i < n; i++)
            a[i] = Long.parseLong(str[i]);
        return a;
    }

    public static String[] getArray() throws IOException {
        return br.readLine().split(" ");
    }

    public static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    public static int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    public static long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }

    public static int lcm(int a, int b) {
        return a * b / gcd(a, b);
    }

    public static int sumDigits(int no) {
        return no == 0 ? 0 : no % 10 + sumDigits(no / 10);
    }

    public static long sumDigits(long no) {
        return no == 0 ? 0 : no % 10 + sumDigits(no / 10);
    }

    public static void reverse(int[] a, int i, int j) {
        while (i < j) {
            a[j] = swap(a[i], a[i++] = a[j--]);
        }
    }

    public static <E> E swap(E a, E b) {
        return a;
    }

    public static void main(String[] args) throws NumberFormatException, IOException {

        int t = nextInt();
        out: while (t-- > 0) {
            int n = nextInt();
            String a = next();
            String b = next();
            if(a.equals(b)) {
                System.out.println("YES");
                continue out;
            }
            int a0 = 0, a1 = 0;
            int b0 = 0, b1 = 0;
            for(int i = 0; i < n; i++) {
                if(a.charAt(i) == '0') a0++;
                else a1++;
                if(b.charAt(i) == '0') b0++;
                else b1++;
            }
            if(a0 != b0 || a1 != b1) {
                System.out.println("NO");
                continue out;
            }
            int k = a.length()-1;
            boolean f = true;
            while(k >= 0) {
                if(f && a.charAt(k) == b.charAt(k)) {
                    if(a.charAt(k) == '0') a0--;
                    else a1--;
                    k--;
                } else if(!f && a.charAt(k) != b.charAt(k)) {
                    if(a.charAt(k) == '0') a1--;
                    else a0--;
                    k--;
                } else {
                    if(a0 != a1) {
                        System.out.println("NO");
                        continue out;
                    }
                    f = !f;
                }
            }
            System.out.println("YES");
        }
    }
}