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.*;
import java.math.*;

public class Main {
	static PrintWriter pw = new PrintWriter(System.out);

	public static void main(String[] args) throws IOException, InterruptedException {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			char[] a = sc.next().toCharArray();
			char[] b = sc.next().toCharArray();
			int[] c0 = new int[a.length], c1 = new int[a.length];
			for (int i = 0; i < a.length; i++) {
				if (a[i] == '0') {
					c0[i] = (i == 0 ? 0 : c0[i - 1]) + 1;
					c1[i] = (i == 0 ? 0 : c1[i - 1]);
				} else {
					c0[i] = (i == 0 ? 0 : c0[i - 1]);
					c1[i] = (i == 0 ? 0 : c1[i - 1]) + 1;
				}
			}
			boolean flip = false, can = true;
//			pw.println(Arrays.toString(c0));
//			pw.println(Arrays.toString(c1));
			for (int i = a.length - 1; i >= 0; i--) {
//				pw.println(i + " " + flip);
				if (a[i] == b[i] && flip) {
					if (c0[i] == c1[i]) {
						flip = false;
					} else {
						can = false;
						break;
					}
				} else if (a[i] != b[i] && !flip) {
					if (c0[i] == c1[i]) {
						flip = true;
					} else {
						can = false;
						break;
					}
				}
			}
			pw.println(can ? "YES" : "NO");
		}
		pw.close();
	}

	public static class tuble implements Comparable<tuble> {
		int x;
		int y;
		int z;

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

		public String toString() {
			return x + " " + y + " " + z;
		}

		public int compareTo(tuble other) {
			if (this.x == other.x) {
				if (this.y == other.y)
					return this.z - other.z;
				return this.y - other.y;
			} else {
				return this.x - other.x;
			}
		}
	}

	public static class pair implements Comparable<pair> {
		int x;
		int y;

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

		public String toString() {
			return x + " " + y;
		}

		public boolean equals(Object o) {
			if (o instanceof pair) {
				pair p = (pair) o;
				return p.x == x && p.y == y;
			}
			return false;
		}

		public int hashCode() {
			return new Integer(x).hashCode() * 31 + new Integer(y).hashCode();
		}

		public int compareTo(pair other) {
			if (this.x == other.x) {
				return Long.compare(this.y, other.y);
			}
			return Long.compare(this.x, other.x);
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader r) {
			br = new BufferedReader(r);
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

	}

}