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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class Solve {

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner();
		PrintWriter pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			String str1 = sc.next();
			String str2 = sc.next();
			if (str1.equals(str2)) {
				System.out.println("YES");
				continue;
			}

			boolean can = true;
			int one1 = 0;
			int zero1 = 0;

			boolean eq = str1.charAt(0) == str2.charAt(0);

			int lastidx = -1;
			for (int i = 0; i < n; i++) {
				if (str1.charAt(i) == '1')
					one1++;
				else
					zero1++;
				if (one1 == zero1) {
					lastidx = i;
				}

			}

			one1 = zero1 = 0;

			for (int i = 0; i <= lastidx; i++) {
				if (str1.charAt(i) == '1')
					one1++;
				else
					zero1++;

				if (eq) {
					if (str1.charAt(i) != str2.charAt(i))
						can = false;
				} else {

					if (str1.charAt(i) == str2.charAt(i))
						can = false;
				}

				if (one1 == zero1) {
					if (i + 1 < n) {
						eq = str1.charAt(i + 1) == str2.charAt(i + 1);

					}
				}

			}

			for (int i = lastidx + 1; i < n; i++) {

				if (str1.charAt(i) != str2.charAt(i))
					can = false;

			}

			System.out.println(can ? "YES" : "NO");

		}

	}

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

		return gcd(b, a % b);

	}

	public static int maxProfit(int[] prices) {
		int n = prices.length;
		int max[] = new int[n];
		max[n - 1] = prices[n - 1];
		for (int i = n - 2; i >= 0; i--) {
			max[i] = Math.max(max[i + 1], prices[i]);
		}

		int ans = 0;
		for (int i = 0; i < n - 1; i++) {
			ans = Math.max(ans, Math.max(0, max[i + 1] - prices[i]));
		}
		return ans;

	}

	private static int sum(long n) {
		int sum = 0;
		while (n > 0) {
			sum += n % 10;
			n /= 10;
		}
		// TODO Auto-generated method stub
		return sum;
	}

	static long nC2(long n) {
		return n * (n - 1) / 2;
	}

	static boolean isPalindrome(String str) {

		// Pointers pointing to the beginning
		// and the end of the string
		int i = 0, j = str.length() - 1;

		// While there are characters toc compare
		while (i < j) {

			// If there is a mismatch
			if (str.charAt(i) != str.charAt(j))
				return false;

			// Increment first pointer and
			// decrement the other
			i++;
			j--;
		}

		// Given string is a palindrome
		return true;
	}

	static long nCrModp(int n, int r, int p) {
		if (r > n - r)
			r = n - r;

		// The array C is going to store last
		// row of pascal triangle at the end.
		// And last entry of last row is nCr
		int C[] = new int[r + 1];

		C[0] = 1; // Top row of Pascal Triangle

		// One by constructs remaining rows of Pascal
		// Triangle from top to bottom
		for (int i = 1; i <= n; i++) {

			// Fill entries of current row using previous
			// row values
			for (int j = Math.min(i, r); j > 0; j--)

				// nCj = (n-1)Cj + (n-1)C(j-1);
				C[j] = (C[j] + C[j - 1]) % p;
		}
		return C[r];
	}

	static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}

}

class Scanner {
	BufferedReader br;
	StringTokenizer st;

	Scanner() {
		br = new BufferedReader(new InputStreamReader(System.in));
	}

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

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

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

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

//creating a hash table
//Hashtable<Integer, String> h = new Hashtable<Integer, String>();
//
//Hashtable<Integer, String> h1 = new Hashtable<Integer, String>();
//
//h.put(3, "Geeks");
//h.put(2, "forGeeks");
//h.put(1, "isBest");
//System.out.println(h.get(1));
//if (h.containsKey(1)) {
//	System.out.println(1111111);
//}
//if (h.containsValue("isBest")) {
//	System.out.println(1111111);
//}
//
//// create a clone or shallow copy of hash table h
//h1 = (Hashtable<Integer, String>) h.clone();
//
//// checking clone h1
////System.out.println("values in clone: " + h1);
//
//// clear hash table h
//h.clear();

//Creating object of the 
// class linked list 
//LinkedList<String> ll 
//    = new LinkedList<String>(); 
//
//// Adding elements to the linked list 
//ll.add("A"); 
//ll.add("B"); 
//ll.addLast("C"); 
//ll.addFirst("D"); 
//ll.add(2, "E"); 
//
//System.out.println(ll); 
//
//ll.remove("B"); 
//ll.remove(3); 
//ll.removeFirst(); 
//ll.removeLast(); 
//
//System.out.println(ll); 

//Set<Integer> set = new HashSet<Integer>();
//try {
//   for(int i = 0; i < 5; i++) {
//      set.add(count[i]);
//   }
//   System.out.println(set);
//
//   TreeSet sortedSet = new TreeSet<Integer>(set);
//   System.out.println("The sorted list is:");
//   System.out.println(sortedSet);
//
//   System.out.println("The First element of the set is: "+ (Integer)sortedSet.first());
//   System.out.println("The last element of the set is: "+ (Integer)sortedSet.last());
//}
//catch(Exception e) {}