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.*;

public class B {
    public static boolean solve(String a, String b){
        if(a.length() != b.length())return false;
        if(Same(a, b, 0, a.length() - 1))return true;
        ArrayList<Integer> lt = new ArrayList<>();
        lt.add(-1);
        int n = a.length();
        int count1 = 0;
        int count2 = 0;
        for(int i = 0; i < n; i++){
            if(a.charAt(i) == '1')count1++;
            else count2++;

            if(count1 == count2)lt.add(i);
        }
//        System.out.println(lt);
        if(lt.size() == 1)return false;
        int size = lt.size();
        if(!Same(a, b, lt.get(size- 1) + 1, n - 1))return false;
        
        
        for(int i = 0; i < size - 1; i++){
            if(!Same(a, b, lt.get(i) + 1, lt.get(i + 1))  && !CompleatlyDiff(a, b, lt.get(i) + 1, lt.get(i + 1)))return false;
        }

        return true;

    }

    public static boolean Same(String a, String b, int i, int n){
        for(; i <= n; i++){
            if(a.charAt(i) != b.charAt(i))return false;
        }
        return true;
    }

    public static boolean CompleatlyDiff(String a, String b, int i, int n){
        for(; i <= n; i++){
            if(a.charAt(i) == b.charAt(i))return false;
        }
        return true;
    }
    
    	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		FastReader sc=new FastReader();
        int t = sc.nextInt();
        while(t-- != 0){
        	int a = sc.nextInt();
        	String x = sc.next();
        	String y = sc.next();
            if(solve(x, y))System.out.println("YES");
            else System.out.println("NO");
        }
	}
		static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
}