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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.text.DecimalFormat;
import java.util.*;

public class Codeforces {
	static int mod=1000_000_007;
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastScanner fs=new FastScanner();
		int t=fs.nextInt();
		while(t-->0) {
			int n=fs.nextInt();
			String a=fs.next()+'0',b=fs.next()+'0';
			solve(n,a.toCharArray(),b.toCharArray());
		}
	}
	static void solve(int n,char a[],char b[]) {
		int cnt=0;
		for(int i=0;i<n;i++) {
			if(a[i]=='0') cnt--;
			else cnt++;
			if((a[i]==b[i])!=(a[i+1]==b[i+1])&&cnt!=0) {
				System.out.println("NO");
				return ;
			}
		}
		System.out.println("YES");
	}
	static void mysort(int[] a) {
		//suffle
		int n=a.length;
		Random r=new Random();
		for (int i=0; i<a.length; i++) {
			int oi=r.nextInt(n), temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		
		//then sort
		Arrays.sort(a);
	}
	
	// Use this to input code since it is faster than a Scanner
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
 
}