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 graphs;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class PW {
	
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader sc = new FastReader();
		int t1=sc.nextInt();
		while(t1-- >0){
		
			int n=sc.nextInt();
			String s=sc.next();
			String t=sc.next();
			int s1 = 0, s2 = 0,flag=0;
			List<Integer> idx=new ArrayList<>();
			 for(int  i = 0 ; i < n ; i ++){
			    s1 += (s.charAt(i)-'0')==1?1:-1;
			    s2 += (s.charAt(i)-'0')==1?1:-1;
			    if(s1 == 0)
			      idx.add(i);
			 }
			 int c = 0;
			 for(int i = n-1 ; i >=0 ; i--) {
			   int ek = s.charAt(i) - '0';
			   int du = t.charAt(i) - '0';
			   if(c==1)
			    ek = 1 - ek;
			   if (ek != du) {
//			     cout << i << endl;
			     if (idx.size() == 0 || idx.get(idx.size()-1) != i) {
			       //System.out.println("NO");
			       flag=1;
			       break;
			     } else {
			       idx.remove(idx.size()-1);
			       c = 1 - c;
			     }
			   }
			   else{
			     if(idx.size()>0 && i==idx.get(idx.size()-1))
			    	 idx.remove(idx.size()-1);
			   }
			 }
			   if(flag==0)
			   System.out.println("YES");
			   else
				   System.out.println("NO");
			}
		
	
	}
	
	
	public static boolean pal(String s)
	{
		int i=0;
		int j=s.length()-1;
		while(i<=j)
		{
			if(s.charAt(i)!=s.charAt(j))
				return false;
			i++;
			j--;
		}
		return true;
	}
	
	
	public static long  solve(long n)
	{
		long sum=0;
		while(true)
		{
			sum=0;
			long t=n;
			while(t!=0)
			{
				sum=sum+t%10;
				t=t/10;
			}
			long ans=gcd(n,sum);
			if(ans>1)
				return n;
			n++;
		}
		   
	}
	
	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 long solve(int n, int r)
    {


        long p = 1, k = 1;
        if (n - r < r) {
            r = n - r;
        }
 
        if (r != 0) {
            while (r > 0) {
                p *= n;
                k *= r;
 
                long m = __gcd(p, k);
 
                
                p /= m;
                k /= m;
 
                n--;
                r--;
            }
 
           
        }
        else {
            p = 1;
        }
 
        
        //System.out.println(p);
        return p;
    }
	
	
	public static long gcd(long a,long b)
	{
		if(a==0||b==0)
			return a+b;
		return gcd(b,(a%b));
	}
	
	public static long __gcd(long n1, long n2)
    {
        long gcd = 1;
 
        for (int i = 1; i <= n1 && i <= n2; ++i) {
            // Checks if i is factor of both integers
            if (n1 % i == 0 && n2 % i == 0) {
                gcd = i;
            }
        }
        return gcd;
    }
	
	public static boolean prime(int n)
	{
		if(n<=2)
			return true;
		
		for(int i=2;i<=Math.sqrt(n);i++)
		{
			if(n%i==0)
				return false;
		}
		
		return true;
		
	}
	public static long fastExpo(long a,long n,long mod){
        if (n == 0)
            return 1;
        else{
            long x = fastExpo(a,n/2,mod);
            if ((n&1) == 1){
                return (((a*x)%mod)*x)%mod;
            }
            else{
                return (((x%mod)*(x%mod))%mod)%mod;
            }
        }
    }
}
class pair{
	//public:
		long f;
		long s;
		pair(long x,long y)
		{
			f=x;
			s=y;
		}
        
}


