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.
//package Codeforces;
import java.util.*;
import java.io.*;
public class B
{
	static final int mod = (int) 1e9 + 7;
	static final int inf = (int) 1e9;
	static final long INF = (long) 1e18;
	public static void main(String[] args) throws IOException {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int T=fs.nextInt();
		while(T-->0)
		{
			int n=fs.nextInt();
		   String str=fs.next();String str2=fs.next();
		   System.out.println(isPos(str,str2) ? "YES":"NO");
		}
	}
	public static boolean isPos(String str,String str2)
	{
		   int diff[]=new int[str.length()];int n=str.length();
		   diff[0]=str.charAt(0)=='1' ? 1:-1;
		   for(int i=1;i<str.length();i++)
		   {
			   diff[i]=diff[i-1];
			   if(str.charAt(i)=='1')diff[i]++;
			   else diff[i]--;
		   }
		   int prev_idx=0;
		   for(int i=0;i<str.length();i++)
		   {
			   if(diff[i]==0)
			   {
				   String str_sub=str.substring(prev_idx,i+1);
				   String str2_sub=str2.substring(prev_idx, i+1);
				   if(!(str_sub.equals(str2_sub) || Inv(str_sub,str2_sub)))
					   return false;
				   prev_idx=i+1;
			   }
		   }
		   String last=str.substring(prev_idx, n);
		   String last2=str2.substring(prev_idx, n);
		   if(!last.equals(last2))
			   return false;
		   if(prev_idx==0 && !str.equals(str2))
			   return false;
		   return true;
	}
	public static boolean Inv(String str,String str2)
	{
		for(int i=0;i<str.length();i++)
		{
			if(str.charAt(i)==str2.charAt(i))return false;
		}
		return true;
	}
	public static int lower_bound(int arr[],int key)
	{
		int index=inf,num=0;
		int left=0,right=arr.length-1;
		while(left<=right)
		{
			int mid=left+(right-left)/2;
			if(arr[mid]>=key) {index=mid;num=arr[mid];right=mid-1;}
			else 
				left=mid+1;
		}
		return index==inf ? arr.length:index;
	}
	public static int upper_bound(int arr[],int key)
	{
		int index=inf,num=0;
		int left=0,right=arr.length-1;
		while(left<=right)
		{
			int mid=left+(right-left)/2;
			if(arr[mid]<=key) {index=mid;num=arr[mid];left=mid+1;}
			else 
				right=mid-1;
		}
		return index==inf ? arr.length:index;
	}
}
class FastScanner {
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer st = new StringTokenizer("");

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

	}

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

	int[] sort(int arr[]) {
		ArrayList<Integer> list = new ArrayList<Integer>();
		for (int i : arr)
			list.add(i);
		Collections.sort(list);
		for (int i = 0; i < arr.length; i++) {
			arr[i] = list.get(i);
		}
		return arr;
	}

	char[] charsort(char arr[]) {
		ArrayList<Character> list = new ArrayList<>();
		for (char c : arr)
			list.add(c);
		Collections.sort(list);
		for (int i = 0; i < list.size(); i++) {
			arr[i] = list.get(i);
		}
		return arr;
	}

	long[] longsort(long arr[]) {
		ArrayList<Long> list = new ArrayList<Long>();
		for (long i : arr)
			list.add(i);
		Collections.sort(list);
		for (int i = 0; i < arr.length; i++) {
			arr[i] = list.get(i);
		}
		return arr;
	}

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

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

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

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