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.
//package codechef;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class A {

	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; 
        } 
    } 
	
//	static boolean isPalindrome(String str)
//    {
// 
//        // Pointers pointing to the beginning
//        // and the end of the string
//        int i = 0, j = str.length() - 1;
// 
//        // While there are characters to compare
//        while (i < j) {
// 
//            // If there is a mismatch
//            if (str.charAt(i) != str.charAt(j))
//                return false;
// 
//            // Increment first pointer and
//            // decrement the other
//            i++;
//            j--;
//        }
// 
//        // Given string is a palindrome
//        return true;
//    }
//    
	
	static void change(char a[],int x)
	{
		for(int i=0;i<=x;i++)
		{
			a[i]=a[i]=='0'?'1':'0';
		}
	}
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader sc=new FastReader();
		int t=sc.nextInt();
		
		while(t-->0)
		{
			int n=sc.nextInt();
			String a=new String();
			a=sc.next();
			String b=new String();
			b=sc.next();
			boolean flag=true;
			boolean check=true;
			int[] cnt=new int[n];
			int ones=0;
			for(int i=0;i<n;i++)
			{
				if(a.charAt(i)=='1')
				{
					ones+=1;
				}
				cnt[i]=ones;
			}
			
			int i=n-1;
			while(i>=0)
			{
				
				
					if (flag) {
						if (a.charAt(i)!=b.charAt(i)) {
							if(cnt[i]*2!=(i+1))
							{
								check=false;
								break;	
							}
							else {
								flag=!flag;
							}
						}
						
						i--;
					}
					else {
						if (a.charAt(i)==b.charAt(i)) {
							if(cnt[i]*2!=(i+1))
							{
								check=false;
								break;	
							}
							else {
								flag=!flag;
							}
						}
						i--;
					}
					
				
			}
			if (check) {
				System.out.println("YES");
			}
			else {
				System.out.println("NO");
			}
		}
	}

}
