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.
/*
*	Author: CrazyDoctor
*	
*	File: B.java
*	
*	Date: 14:52:40, 03.04.2021
*/

import java.io.*;
import java.util.*;

public class B {
	public static void main(String args[]) {
		FastScanner sc = new FastScanner();
		FastPrinter fp = new FastPrinter();
		

		int t = sc.nextInt();

		while(t-->0) {

			int n = sc.nextInt();
			String a = sc.next().trim();
			String b = sc.next().trim();
			int ones = 0;
			int zeroes = 0;
			int[] flags = new int[n];
			boolean yes = true;
			for (int i = 0; i < n; i++) {
				if (a.charAt(i) == '1') {
					ones++;
				} else {
					zeroes++;
				}
				if (ones == zeroes) {
					flags[i] = 1;
				}
			}

			boolean rev = true;
			int i = n-1;
			while(i >= 0) {
				if (rev) {
					if (b.charAt(i) == a.charAt(i)) {
						i--;
					} else {
						if (flags[i] == 1) {
							i--;
							rev = !rev;
						} else {
							yes = false;
							i = -1;
						}
					}
				} else {
					if (b.charAt(i) == (a.charAt(i) == '1' ? '0' : '1')) {
						i--;
					} else {
						if (flags[i] == 1) {
							i--;
							rev = !rev;
						} else {
							yes = false;
							i = -1;
						}
					}
				}
			}
			
			fp.println((yes ? "YES" : "NO"));
		}

		fp.close();
	}

	static class FastPrinter extends PrintWriter {

		FastPrinter() {
			super(System.out);
		}

		void println(int n, int[] a) {
			for (int i = 0; i < n; i++) {
				this.print(a[i] + (i == n - 1 ? "\n" : " "));
			}
		}

		void println(int n, long[] a) {
			for (int i = 0; i < n; i++) {
				this.print(a[i] + (i == n - 1 ? "\n" : " "));
			}
		}

		void println(int n, double[] a) {
			for (int i = 0; i < n; i++) {
				this.print(a[i] + (i == n - 1 ? "\n" : " "));
			}
		}

		void println(ArrayList<Object> a) {
			int n = a.size();
			for (int i = 0; i < n; i++) {
				this.print(a.get(i) + (i == n - 1 ? "\n" : " "));
			}
		}
	}

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		
		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}

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

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

		long[] nextLongA(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = this.nextLong();
			return a;
		}

		double[] nextDoubleA(int n) {
			double[] a = new double[n];
			for (int i = 0; i < n; i++)
				a[i] = this.nextDouble();
			return a;
		}

		ArrayList<Integer> nextIntAL(int n) {
			ArrayList<Integer> a = new ArrayList<Integer>(n);
			for (int i = 0; i < n; i++)
				a.add(this.nextInt());
			return a;
		}

		ArrayList<Long> nextLongAL(int n) {
			ArrayList<Long> a = new ArrayList<Long>(n);
			for (int i = 0; i < n; i++)
				a.add(this.nextLong());
			return a;
		}

		ArrayList<Double> nextDoubleAL(int n) {
			ArrayList<Double> a = new ArrayList<Double>(n);
			for (int i = 0; i < n; i++)
				a.add(this.nextDouble());
			return a;
		}

		ArrayList<String> nextStringAL(int n) {
			ArrayList<String> a = new ArrayList<String>(n);
			for (int i = 0; i < n; i++)
				a.add(this.next());
			return a;
		}
	}
}