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.io.PrintWriter;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.FileReader;
import java.io.IOException;
import java.util.StringTokenizer;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Arrays;
import java.util.Stack;
import java.util.HashSet;
import java.util.TreeSet;
import java.util.TreeMap;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.ArrayDeque;
import static java.lang.Math.min;
import static java.lang.Math.max;
import static java.lang.Math.sqrt;
import static java.lang.Math.pow;
import static java.lang.Math.abs;
import static java.lang.Integer.MIN_VALUE;
import static java.lang.Integer.MAX_VALUE;

 
public class Main { 
    
    static boolean ONLINE_JUDGE = false;
    static Fast f = new Fast();
    static PrintWriter out = new PrintWriter(System.out); 
    static boolean TEST_CASES =  true;
    static int mod1 = (int)1e9+7;
    static int mod2 = 998244353;

    static void solve() {        

      int n = ri();
      String a = rs();
      String b = rs();
      int ind = -1;
      int[][] dp = new int[n][2];
      dp[0][a.charAt(0)-48]++;
      for(int i = 1; i < n; i++){
        dp[i][0] = dp[i-1][0];
        dp[i][1] = dp[i-1][1];
        dp[i][a.charAt(i)-48]++;
      }
      boolean flag = true;
      boolean flag2 = true;
     for(int i = n-1; i >-1; i--) {
      if(flag){
          if(a.charAt(i)==b.charAt(i)) continue;

          if(dp[i][0]==dp[i][1]) {
            flag = false;
          }

          else {
            flag2 = false;
            break;
          }
      }
      else {
        if(a.charAt(i)!=b.charAt(i)) continue;
        if(dp[i][0]==dp[i][1]) {
          flag = true;
        }
         else {
            flag2 = false;
            break;
          }
      }


      
     }


     out.println(flag2?"YES":"NO");








    }
   


    public static void main(String[] args)throws Exception{


      
      
      if(TEST_CASES){
          
          int t = ri();
          
          while(t-->0){
          
            solve();
          
          }

      }

      else {

        solve();
      
      }
      
      out.close();

    }

    
    static int nod(long l) {
       if(l>=1000000000000000000l) return 19;
       if(l>=100000000000000000l) return 18;
       if(l>=10000000000000000l) return 17;
       if(l>=1000000000000000l) return 16;
       if(l>=100000000000000l) return 15;
       if(l>=10000000000000l) return 14;
       if(l>=1000000000000l) return 13;
       if(l>=100000000000l) return 12;
       if(l>=10000000000l) return 11;
       if(l>=1000000000) return 10;
       if(l>=100000000) return 9;
       if(l>=10000000) return 8;
       if(l>=1000000) return 7;
       if(l>=100000) return 6;
       if(l>=10000) return 5;
       if(l>=1000) return 4;
       if(l>=100) return 3;
       if(l>=10) return 2;
       return 1;
    }


    static int ri() {

      return f.nextInt();
    
    }

    static long rl() {
 
      return f.nextLong();
    
    }

    static String rs(){

      return f.next();

    }

    static String rS(){
      
       return f.nextLine();
    
    }

    static char rc(){

         return f.next().charAt(0);

    }

    static int[] ra(int n) {
        
      int[] a = new int[n];
      for(int i = 0;i<n;i++) a[i] = ri();
      return a;
        
    }

    static long[] ral(int n) {
        
      long[] a = new long[n];
      for(int i = 0;i<n;i++) a[i] = rl();
      return a;
        
    }

    static char[] rac(){

        char[] c = rs().toCharArray();

        return c;

    }
    
    static int[][] rm(int n, int m){
       
       int[][] mat = new int[n][m];

       for(int i = 0; i < n; i++) mat[i] = ra(m);
       
       return mat;

    }

    static char[][] rmc(int n){
      
      char[][] cmat = new char[n][];
      
      for(int i = 0; i < n;i++) cmat[i] = rac();

      return cmat;

    }

    static void sort(int[] a) {

      ArrayList<Integer> list=new ArrayList<>();
      for (int i:a) list.add(i);
      Collections.sort(list);
      for (int i=0; i<a.length; i++) a[a.length-1-i]=list.get(i);

    }

    static void sort(double[] a) {

      ArrayList<Double> list=new ArrayList<>();
      for (double i:a) list.add(i);
      Collections.sort(list);
      for (int i=0; i<a.length; i++) a[i]=list.get(i);

    }
  
    static class Fast{
       public BufferedReader br;
       public StringTokenizer st;
       
       public Fast(){
            try{
                br = ONLINE_JUDGE? (new BufferedReader(new FileReader("input.txt"))):(new BufferedReader(new InputStreamReader(System.in)));
            }
            catch(Exception e){
              throw new RuntimeException(e);
            }
       }
       
       String next(){
            while(st==null || !st.hasMoreTokens()){
                 try{
                      st=new StringTokenizer(br.readLine());
                 }
                 catch(IOException e){
                      throw new RuntimeException(e);
                 }
                 
            }
                 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; 
          } 
   
     }
}



 