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 AgainJava;
import java.util.Scanner ;
import java.io.*;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class NewProg {
    public static void main(String[] args) {
		FastScanner fs=new FastScanner();
                int t = fs.nextInt() ;
                while(t --  > 0  ){
                    int n = fs.nextInt(),o=0,z=0;
                    String a = fs.next() , b = fs.next() ; 
                    ArrayList<Integer> vec = new ArrayList<>() ; 
                    vec.add(-1);
                    for(int i =0 ; i < n ; i++){
                        if(a.charAt(i) == '0'){
                            z++;
                        }
                        else{
                            o++; 
                        }
                        if(z == o){
                            // ith index is important 
                            vec.add(i) ; 
                        }
                    }
                    int offset = 0  ;  
                    Collections.reverse(vec);
                    int e = n-1 , cnt = 0 ;
                    Boolean f = true;
                    for(int x : vec){
                        if(cnt == 0){
                            for(int i = x+1 ; i <= e ; i++){
                                if(a.charAt(i) != b.charAt(i)){
                                    f = false ; 
                                }
                            }
                            e = x ;
                            cnt++;
                            continue ; 
                        }
                        Boolean ok1 = false , ok2 = false  ;
                        for(int i = x+1 ; i <= e ; i++){
                            if(a.charAt(i) != b.charAt(i)){
                                    ok1 = true ; 
                            }
                            else{
                                ok2 = true  ;
                            }
                        } 
                        if(ok1 == true && ok2 == true){
                            f = false ;
                            break ; 
                        }
                        else if(ok1 == true){
                            if(offset == 0){
                                offset = 1 ;
                            }
                        }
                        else if(ok2 == true){
                            if(offset == 1){
                                offset = 0 ; 
                            }
                        }
                        e = x ; 
                    }
                    if(f == true){
                        System.out.println("YES");
                    }
                    else{
                        System.out.println("NO");
                    }
                }
    }

    
    static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		public String next() {
			while (!st.hasMoreElements())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int[] readArray(int n) {
			int[] a=new int[n]; 
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		int nextInt () {
			return Integer.parseInt(next());
		}
    }
}