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.Scanner;
import java.util.Arrays;
import java.util.HashMap;
//import java.util.HashMap;
//import java.util.Arrays;
//import java.util.Stack;
import java.lang.Math;
public class solution {
	
	
	                
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int testcases = s.nextInt();
		while(testcases>0) {
			
			int n=s.nextInt();
			String a=s.next();
			String b=s.next();
			int a1=0,a0=0;
			int checkpoint=-1;
			boolean mainans=true;
			
			if(a.equals(b)) {
				System.out.println("YES");
				testcases--;
				continue;
			}
			
			for(int i=0;i<n;i++) {
				if(a.charAt(i)=='1')
					a1++;
				else
					a0++;
				
				if(a1==a0) {
					boolean ans=true;
					if(a.charAt(checkpoint+1)==b.charAt(checkpoint+1)) {
					for(int j=checkpoint+1;j<=i;j++) {
						if(a.charAt(j)!=b.charAt(j)) {
							ans=false;
							break;
						}
					}
					if(!ans) {
						mainans=false;
						break;
					}
					checkpoint=i;
					}else {
						for(int j=checkpoint+1;j<=i;j++) {
							if(a.charAt(j)==b.charAt(j)) {
								ans=false;
								break;
							}
						}
						if(!ans) {
							mainans=false;
							break;
						}
						checkpoint=i;
					}
				}
				
			}
			if(mainans) {
				if(a.substring(checkpoint+1).equals(b.substring(checkpoint+1))) {
				System.out.println("YES");
				}
				else
					System.out.println("NO");
			}else {
				System.out.println("NO");
			}
			
			
				testcases-=1;
			}
		
	}
//	
	

}

