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.*;
import java.util.*;

public class B7122 {
	static int t;
	public static void main(String[] args) throws IOException {
		init_io();
		t = Integer.parseInt(br.readLine());
		while(t-- > 0) {
			int len = Integer.parseInt(br.readLine());
			String line1 = br.readLine(), line2 = br.readLine();
			boolean[] canFlip = new boolean[len];
			int cnt0 = 0, cnt1 = 0;
			for (int i = 0; i < len; i++) {
				if (line1.charAt(i) == '0') {
					cnt0++;
				}
				else {
					cnt1++;
				}
				if (cnt0 == cnt1) {
					canFlip[i] = true;
				}
			}
			boolean fail = false;
			int flips = 0;
			for (int i = len-1; i >= 0; i--) {
				if (flips % 2 > 0) {
					if (line1.charAt(i) != line2.charAt(i)) continue;
					else if (canFlip[i]) {
						flips++;
					}
					else {
						fail = true;
						break;
					}
				}
				else {
					if (line1.charAt(i) == line2.charAt(i)) continue;
					else if (canFlip[i]) {
						flips++;
					}
					else {
						fail = true;
						break;
					}
				}
			}
			if (fail) {
				out.println("NO");
			}
			else {
				out.println("YES");
			}
		}
		out.close();
	}

	static StreamTokenizer in;
	static PrintWriter out;
	static BufferedReader br;

	static int nint() throws IOException {
		in.nextToken();
		return (int) in.nval;
	}

	static void init_io() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		in = new StreamTokenizer(br);
		out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	}
}