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 Main
 
{
    class Pair<S extends Comparable<S>, T extends Comparable<T>> implements Comparable<Pair<S, T>> {
        S first;
        T second;
    
        Pair(S f, T s) {
            first = f;
            second = s;
        }

        @Override
        public int compareTo(Pair<S, T> o) {
            int t = first.compareTo(o.first);
            if (t == 0) return second.compareTo(o.second);
            return t;
        }
    
        @Override
        public int hashCode() {
            return (31 + first.hashCode()) * 31 + second.hashCode();
        }
    
        @Override
        public boolean equals(Object o) {
            if (!(o instanceof Pair)) return false;
            if (o == this) return true;
            Pair p = (Pair) o;
            return first.equals(p.first) && second.equals(p.second);
        }
    
        @Override
        public String toString() {
            return "Pair{" + first + ", " + second + "}";
        }
    }
    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; 
        } 
    } 
    public static void main (String[] args) throws java.lang.Exception
    {
        // try {
        //     System.setIn(new FileInputStream("input.txt"));
        //     System.setOut(new PrintStream(new FileOutputStream("output.txt")));
        // } catch (Exception e) {
        //     System.err.println("Error");
        // }
        
        FastReader sc = new FastReader();
        int t =sc.nextInt();
        PrintWriter out=new PrintWriter(System.out);
        while(t-->0){
            solve(sc,out);
        }
         out.close();
    }
    static void solve(FastReader sc,PrintWriter out) {
        int n = sc.nextInt();
        char a[] = sc.next().toCharArray();
        char b[] = sc.next().toCharArray();
        int cnt0=0,cnt1=0;
        boolean arr[] = new boolean[n];
        for(int i=0;i<n;i++){
            if(a[i]=='0')cnt0++;
            else cnt1++;
            if(i%2!=0 && (cnt0==cnt1)){
                arr[i]=true;
            }

        }
        int step=0;
        for(int i=n-1;i>=0;i--){
            if(step%2!=0){
                if(a[i]=='0'){
                    a[i]='1';
                }
                else{
                    a[i]='0';
                }
            }
            if(a[i]!=b[i]){
                if(!arr[i]){
                    out.println("NO");
                    return;
                }
                else{
                    step++;

                }
            }
        }
        out.println("YES");
        return;



    }
}