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.lang.*;
import java.io.*;
public class  a  {
    static final int mod = 1000000007; 
    static FastReader sc = new FastReader();  
    static PrintWriter out = new PrintWriter(System.out);
        
    public static void main (String[] args) throws java.lang.Exception {
        int t = sc.nextInt();
        while (t-->0) {
            int n = sc.nextInt();
            String first = sc.nextLine();
            String second = sc.nextLine();
            int fcnt = 0 ,scnt = 0;
            int farr[] = new int[n];
            int sarr[] = new int[n];
            for ( int i = 0 ; i < n; i++ ) {
                if (first.charAt(i) == '1')
                    fcnt++;
                if (second.charAt(i) == '1')
                    scnt++;
                farr[i] = fcnt;
                sarr[i] = scnt;
            }
            if ( n % 2 != 0  && first.charAt(n-1) != second.charAt(n-1)) {
                out.println("NO");
                continue;
            }
            if(n % 2 != 0 )
                n -= 1;
            boolean flag = true , tple = false;
            for ( int i = n-1 ;i  >= 0 ; i -= 2) {
                int fct = farr[i];
                int sct = sarr[i];
                char a = first.charAt(i);
                char b = first.charAt(i-1);
                if (tple) {
                        fct = i+1-fct;
                        sct = i+1-sct;
                        if(a =='1')
                            a = '0';
                        else
                            a = '1';
                        if(b =='0')
                            b = '1';
                        else
                            b = '0';
                }
                if(a != second.charAt(i) && b != second.charAt(i-1)) {
                                        
                    if(2*fct != i+1) {
                        flag = false;
                        break;
                    }else 
                        tple = (!tple);
                }else if (a == second.charAt(i) && b == second.charAt(i-1)){
                    continue;
                }else {
                    flag = false;
                    break;
                }
            }
            if (flag)
                out.println("YES");
            else
                out.println("NO");
        }
        out.flush();        
    }
}
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; 
        } 
    }