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.*;
import java.io.*;
public class Main {
	static BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
	static BufferedWriter wr = new BufferedWriter(new OutputStreamWriter(System.out));
	static StringTokenizer tok;
	public static void main(String[] args) throws Exception {
		solution2();
	}
	
	public static void solution2() throws Exception {
		int TestCase = Integer.parseInt(rd.readLine());
		for(int TT=0;TT<TestCase;TT++) {
			int n = Integer.parseInt(rd.readLine());
			char[] a = rd.readLine().toCharArray();
			char[] b = rd.readLine().toCharArray();
			
			boolean start = false;
			boolean chgau = true;
			boolean pass = true;
			int[] cnt = new int[2];
			for(int i=n-1;i>=0;i--) {
				if(!start && a[i] == b[i]) continue;	// 끝의값부터 이미 같은 값들은 패스한다.
				else if(!start) start = true;
				
				if(chgau) {
					if(cnt[0] != cnt[1] && a[i] == b[i]) {
						pass = false;
						break;	// 0의 개수와 1의 개수가 같지 않은데 같은값이 나왔으면 NO
					}
					else if(a[i] == b[i]) {
						chgau = false;
//						wr.write(Arrays.toString(a)+" "+Arrays.toString(cnt) +" "+i);
//						wr.newLine();
						i++;
						cnt[0] = 0;
						cnt[1] = 0;
						continue;
					} // 0의 개수와 1의 개수가 같으며 같은값이 나오면 다시 돌려서 같은 값을 체크하도록 한다.
					else {
						cnt[a[i]-'0']++;
					}	// 그냥 다른값이 나온 경우 카운트를 올려준다.
				} else {
					if(cnt[0] != cnt[1] && a[i] != b[i]) {
						pass = false;
						break;	// 0의 개수와 1의 개수가 같지 않은데 다른값이 나왔으면 NO
					}
					else if(a[i] != b[i]) {
						chgau = true;
//						wr.write(Arrays.toString(a)+" "+Arrays.toString(cnt) +" "+i);
//						wr.newLine();
						i++;
						cnt[0] = 0;
						cnt[1] = 0;
						continue;
					} // 0의 개수와 1의 개수가 같으며 다른값이 나오면 다시 돌려서 같은 값을 체크하도록 한다.
					else {
						cnt[a[i]-'0']++;
					}	// 그냥 같은 값이 나온 경우 카운트를 올려준다.
				}
				
				if(i == 0 && cnt[0] != cnt[1]) pass = false;		// 맨앞에서 또한 카운트의 개수가 같지 않으면 NO
			}
			wr.write(pass ? "YES" : "NO");
			wr.newLine();
		}
		wr.flush();
	}
	
	public static void solution() throws Exception {
		int TestCase = Integer.parseInt(rd.readLine());
		for(int TT=0;TT<TestCase;TT++) {
			StringBuilder str = new StringBuilder(rd.readLine());
			int n = str.length();
			boolean yes = false;
			for(int i=0;i<n;i++) {
				if(str.charAt(i) != 'a') {
					str.insert(n-i, 'a');
					yes = true;
					break;
				}
			}
			if(yes) {
				System.out.println("YES");
				System.out.println(str);
			} else System.out.println("NO");
		}
	}
}