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.
// Main Code at the Bottom
import java.util.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.io.*; 
public class Main{
	//Fast IO class
    static class FastReader {
        BufferedReader br; 
        StringTokenizer st; 
        public FastReader() {
        	boolean env=System.getProperty("ONLINE_JUDGE") != null;
        	if(!env) {
        		try {
					br=new BufferedReader(new FileReader("input.txt"));
				} catch (FileNotFoundException e) {
					e.printStackTrace();
				}
        	}
        	else 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; 
        } 
    }     
    static long MOD=(long)1e9+7;
    //debug
    static void debug(Object... o) {
        System.out.println(Arrays.deepToString(o));
    }
    static FastReader sc=new FastReader();
    static PrintWriter out=new PrintWriter(System.out);  
    //Global variables and functions
    //Main function(The main code starts from here)
    public static void main (String[] args) throws java.lang.Exception {
    	int test=1;
    	test=sc.nextInt();
    	while(test-->0) {
    		int n=sc.nextInt();
    		char s[]=sc.nextLine().toCharArray();
    		char t[]=sc.nextLine().toCharArray();
    		int z1=0,z2=0;
    		String ans ="YES";
    		for(int i=0;i<n;i++) {
    			if(s[i]=='0') z1++;
    			if(t[i]=='0') z2++;
    			if(z1==z2 || z1==i+1-z2) continue;
    			ans="NO";
    		}
    		if(z1!=z2) ans = "NO";
    		out.println(ans);
    		
    	}
        out.flush();
        out.close();
    }
}