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

public class B {

    public static void main(String[] arg) {
        int testCase;
        FastScanner s = new FastScanner();
        testCase = s.nextInt();

        while (testCase > 0) {
            int len = s.nextInt();
            String str1 = s.next();
            String str2 = s.next();
            int[] ones = new int[len];
            int[] zeros = new int[len];
            if(str1.charAt(0)=='1'){
                ones[0] = 1;
                zeros[0] = 0;
            }
            else{
                ones[0] = 0;
                zeros[0] = 1;
            }
            for(int i = 1; i < len; i++){
                ones[i] = ones[i-1];
                zeros[i] = zeros[i-1];
                if(str1.charAt(i) == '1'){
                    ones[i]++;
                }
                else{
                    zeros[i]++;
                }
            }

            boolean canNotDo = false;
            int numOfOperationTillNow = 0;
            for(int i = len - 1; i >= 0; i--){
                if(str1.charAt(i) != str2.charAt(i)){
                    if(numOfOperationTillNow % 2 != 1){
                        if(zeros[i] == ones[i]){
                            numOfOperationTillNow++;
                        }
                        else{
                            canNotDo = true;
                            break;
                        }
                    }
                }
                else{
                    if(numOfOperationTillNow % 2 != 0){
                        if(zeros[i] == ones[i]){
                            numOfOperationTillNow++;
                        }
                        else{
                            canNotDo = true;
                            break;
                        }
                    }
                }
            }

            if(canNotDo){
                System.out.println("NO");
            }
            else{
                System.out.println("YES");
            }
            testCase--;
        }
        
    }
    public static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }

        int nextInt () {
            return Integer.parseInt(next());
        }
    }

}
