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.
/*input
6
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
8
10010111
10101011

*/
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
	static PrintWriter out;
	static int MOD = 1000000007;
	static FastReader scan;

    /*-------- I/O using short named function ---------*/
	public static String ns(){return scan.next();}
	public static int ni(){return scan.nextInt();}
	public static long nl(){return scan.nextLong();}
	public static double nd(){return scan.nextDouble();}
	public static String nln(){return scan.nextLine();}
	public static void p(Object o){out.print(o);}
	public static void ps(Object o){out.print(o + " ");}
	public static void pn(Object o){out.println(o);}
    /*-------- for output of an array ---------------------*/
     static void iPA(int arr []){
     	 StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void lPA(long arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void sPA(String arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void dPA(double arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
    /*-------------- for input in an array ---------------------*/
	static void iIA(int arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = ni();
	}
	static void lIA(long arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = nl(); 
	}
	static void sIA(String arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = ns(); 
	}
	static void dIA(double arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = nd();
	}
	/*------------ for taking input faster ----------------*/
     static class FastReader
	{
		BufferedReader br;
		StringTokenizer st;

		public FastReader()
		{
			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());
		}

		String nextLine()
		{
			String str = "";
			try
			{
				str = br.readLine();
			}
			catch (IOException e)
			{
				e.printStackTrace();
			}
			return str;
		}
	}

	// Method to check if x is power of 2
    static boolean isPowerOfTwo (int x) { return x!=0 && ((x&(x-1)) == 0);}
    //Method to return lcm of two numbers
    static int gcd(int a, int b){return a==0?b:gcd(b % a, a); }
    //Method to count digit of a number
    static int countDigit(long n){return (int)Math.floor(Math.log10(n) + 1);}

    //Method for sorting
	static void ruffle_sort(int[] a) {
		//shandom_ruffle
		Random r=new Random();
		int n=a.length;
		for (int i=0; i<n; i++) {
			int oi=r.nextInt(n);
			int temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		//sort
		Arrays.sort(a);
	}
	//Method for checking if a number is prime or not
	static boolean isPrime(int n) { 
        if (n <= 1) return false; 
        if (n <= 3) return true; 
        if (n % 2 == 0 || n % 3 == 0) return false; 
        for (int i = 5; i * i <= n; i = i + 6) 
            if (n % i == 0 || n % (i + 2) == 0) 
            return false; 
        return true; 
    } 

	public static void main (String[] args) throws java.lang.Exception
	{
		OutputStream outputStream =System.out;
        out =new PrintWriter(outputStream);
        scan =new FastReader();
		//for fast output sometimes
		StringBuilder sb = new StringBuilder();
		int t = ni();
		while(t-->0){
			int n = ni();
			String s1 = ns(), s2 = ns();
			char c1[] = s1.toCharArray();
			char c2[] = s2.toCharArray();
			int cnt[] = new int[n];
			boolean equal = true;
			int one[] = new int[n];
			int count = 0;
			if(c1[n-1]!=c2[n-1]){
				equal = false;
				count=1;
				cnt[n-1]=1;
			}
			for(int i=n-2; i>=0; i--){
				if(equal){
					if(c1[i]!=c2[i]){
						count++;
						equal=false;
					}
				}else{
					if(c1[i]==c2[i]){
						count++;
						equal = true;
					}
				}
				cnt[i]=count;
			}
			int sum = 1;
			int o=0;
			for(int i=0; i<n;i++){
				if(c1[i]=='1')
					o++;
				else
					o--;
				one[i]=o;
			}
			boolean check = true;
			for(int i=1; i<n; i++){
				if(cnt[i]!=cnt[i-1]){
					if(one[i-1]!=0){
						check = false;
						break;
					}
				}
			}
			if(cnt[n-1]!=0){
				if(one[n-1]!=0){
					check = false;
				}
			}
			//iPA(cnt);
			//iPA(one);
			if(check){
				pn("YES");
			}else{
				pn("NO");
			}
        }
        out.flush();
		out.close();
	}
	
}

