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;

public class FlipTheBits {

	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int T = Integer.parseInt(in.readLine());
		for (int t = 0; t < T; t++) {
			int N = Integer.parseInt(in.readLine());
			char[] fromCS = in.readLine().toCharArray();
			char[] toCS = in.readLine().toCharArray();

			boolean[] from = convert(fromCS);
			boolean[] to = convert(toCS);

			int[] ones = new int[N];
			ones[0] = fromCS[0] - '0';
			for (int i = 1; i < from.length; i++) {
				ones[i] = ones[i - 1] + (fromCS[i] - '0');
			}

			boolean answer = true;

			boolean cur = false;
			for (int i = N - 1; i >= 0; i--) {
				if (from[i] ^ cur != to[i]) {
					if (i % 2 == 0) {
						answer = false;
						break;
					} else if (ones[i] == (i + 1) / 2) {
						cur = !cur;
					} else {
						answer = false;
						break;
					}
				}
			}

			System.out.println((answer) ? "YES" : "NO");

		}
	}

	public static boolean[] convert(char[] cs) {
		boolean[] bs = new boolean[cs.length];
		for (int i = 0; i < bs.length; i++) {
			bs[i] = cs[i] == '1';
		}
		return bs;
	}

}
