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.Scanner;

public class flipthebits {

	public static void main(String[] args) {

		Scanner scn = new Scanner(System.in);

		int t = scn.nextInt();
		for (int tt = 0; tt < t; tt++) {
			int n = scn.nextInt();
			String a = scn.next();
			String b = scn.next();
			int count0 = 0;
			int count1 = 0;
			int[] ones = new int[n];
			int[] zeros = new int[n];
			if (a.charAt(0) == '0') {
				zeros[0] = 1;
			} else {
				ones[0] = 1;
			}
			for (int i = 1; i < n; i++) {
				if (a.charAt(i) == '0') {
					zeros[i] = zeros[i - 1] + 1;
					ones[i] = ones[i - 1];
				} else {
					zeros[i] = zeros[i - 1];
					ones[i] = ones[i - 1] + 1;
				}
			}
			boolean enter = false;
			boolean flag = true;
			if (a.charAt(n - 1) == b.charAt(n - 1)) {
				if (zeros[n-1] == ones[n-1]) {
					flag = true;
				}
			} else {
				if (zeros[n-1] == ones[n-1]) {
					flag = false;
				}else {
					System.out.println("NO");
					continue;
				}
			}
			for (int i = n - 2; i >= 0; i--) {
				if (flag) {
					if (a.charAt(i) != b.charAt(i)) {
						if (zeros[i] == ones[i]) {
							flag = false;
						} else {
							System.out.println("NO");
							enter = true;
							break;
						}
					}
				} else {
					if (a.charAt(i) == b.charAt(i)) {
						if (zeros[i] == ones[i]) {
							flag = true;
						} else {
							System.out.println("NO");
							enter = true;
							break;
						}
					}
				}
			}
			if (!enter) {
				System.out.println("YES");
			}
		}

	}

}
