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 Template {
	static int n;
	static int[] zeros = new int[(int)(3*1e5+1)];
	static int[] ones = new int[(int)(3*1e5+1)];
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int yo = sc.nextInt();
		outer: while (yo-- > 0) {
			n = sc.nextInt();
			char[] a = sc.next().toCharArray();
			char[] b = sc.next().toCharArray();
			for(int i =0; i < n; i++) {
				if(a[i] == '0') {
					zeros[i] = i-1 < 0 ? 1 : (1+zeros[i-1]);
					ones[i] = i-1 < 0 ? 0 : ones[i-1];
				}
				else {
					zeros[i] = i-1 < 0 ? 0 : zeros[i-1];
					ones[i] = i-1 < 0 ? 1 : (1+ones[i-1]);
				}
			}
			
		
			boolean flip = false;
			for(int i = n-1; i >= 0; i--) {
				if(flip) {
					a[i] = a[i] == '0' ? '1' : '0';
				}
				if(a[i] == b[i]) {
					continue;
				}
				if(a[i] != b[i]) {
					int z = zeros[i];
					int o = ones[i];
					if(z == o) {
						flip = !flip;
					}
					else {
						out.println("No");
						continue outer;
					}
				}
			}
			out.println("Yes");
		}
		out.close();
	}

	static class Pair {
		int x;
		int y;

		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
	}

	static void ruffleSort(int[] a) {
		int n = a.length;
		Random r = new Random();
		for (int i = 0; i < a.length; i++) {
			int oi = r.nextInt(n), temp = a[i];
			a[i] = a[oi];
			a[oi] = temp;
		}
		Arrays.sort(a);
	}

	static long gcd(long a, long b) {
		if (b == 0)
			return a;
		return gcd(b, a % b);
	}

	static boolean[] sieve(int n) {
		boolean isPrime[] = new boolean[n + 1];
		for (int i = 2; i <= n; i++) {
			if (isPrime[i])
				continue;
			for (int j = 2 * i; j <= n; j += i) {
				isPrime[j] = true;
			}
		}
		return isPrime;
	}

	static int mod = 1000000007;

	static long pow(int a, long b) {
		if (b == 0) {
			return 1;
		}
		if (b == 1) {
			return a;
		}
		if (b % 2 == 0) {
			long ans = pow(a, b / 2);
			return ans * ans;
		} else {
			long ans = pow(a, (b - 1) / 2);
			return a * ans * ans;
		}

	}

	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

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

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

		long nextLong() {
			return Long.parseLong(next());
		}
	}

	//	For Input.txt and Output.txt	
	//	FileInputStream in = new FileInputStream("input.txt");
	//	FileOutputStream out = new FileOutputStream("output.txt");
	//	PrintWriter pw = new PrintWriter(out);
	//	Scanner sc = new Scanner(in);
}
