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.

//..............Learning.........//

import java.util.*;
import java.io.*;
import java.lang.*;
public class B {

static FastReader  sc  = new FastReader();
static PrintWriter out = new PrintWriter(System.out);

public static void main(String[] args) throws Exception {
// try {
//	sieve();
//Arrays.fill(dp, -1);
  int t = sc.nextInt();
  while (t-- > 0)
  B.go();
  out.flush();
// } catch (Exception e) {
//    return;}
}
// >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Code Starts <<<<<<<<<<<<<<<<<<<<<<<<<<<<< //
static int mod=998244353;
static HashSet<Integer> set=new HashSet<>();
static void go() {
int n=sc.nextInt();
String a=sc.next();
String b=sc.next();
int pos=0;
//for(int i=0;i<n;i++) {
//	if(a.charAt(i)!=b.charAt(i)) {
//		pos=i;
//		break;
//	}
//}
int pos2=n-1;
//for(int i=n-1;i>=0;i--) {
//	if(a.charAt(i)!=b.charAt(i)) {
//		pos2=i;
//		break;
//	}
//}
int zero[]=new int[n];

int one[]=new int[n];
int aa[]=new int[n];
int bb[]=new int[n];
for(int i=0;i<n;i++) {
	if(a.charAt(i)=='0') {
		aa[i]=0;
	}else {
		aa[i]=1;
	}
	if(b.charAt(i)=='0') {
		bb[i]=0;
	}else {
		bb[i]=1;
	}
	if(i==pos) {
		if(a.charAt(i)=='0') {
			zero[i]=1;
		}
	}else {
		if(a.charAt(i)=='0') {
			zero[i]=zero[i-1]+1;
		}else
			zero[i]=zero[i-1];
	}
	if(i==pos) {
		if(a.charAt(i)=='1') {
			one[i]=1;
		}
	}else {
		if(a.charAt(i)=='1') {
			one[i]=one[i-1]+1;
		}else
			one[i]=one[i-1];
	}
}

//out.println(Arrays.toString(zero)+" "+Arrays.toString(one));
for(int i=n-1;i>=0;i--) {
	
	if(aa[i]!=bb[i]) {
	if(zero[i]==one[i]) {

			for(int j=i;j>=0;j--) {
				if(i!=j) {
					if(zero[j]==one[j]) {
						break;
					}
				}
				aa[j]=1-aa[j];
			}
	}else {
		out.println("NO");
		return;
	}
	}
}
for(int i=0;i<n;i++) {
	if(aa[i]!=bb[i]) {
		out.println("NO");
		return;
	}
}
//for(int i=pos;i<=pos2;i++) {
//
//		if(zero[i]!=one[i]) {
//		out.println("NO");
//		return;
//		}
//	
//}

out.println("YES");
}


// >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Code Ends <<<<<<<<<<<<<<<<<<<<<<<<<<<<< //


static long gcd(long a,long b){
    if(b==0){
        return a;
    }return gcd(b,a%b);
}
static long fact[];
static long invfact[];
static long ncr(int n,int k) {
	if(k<0||k>n) {
		return 0;
	}
	long x=fact[n]%mod;
	long y=invfact[k]%mod;
	long yy=invfact[n-k]%mod;
	long ans=(x*y)%mod;
	ans=(ans*yy)%mod;
	return ans;
}
static int prime[]=new int[1000006];
static void sieve() {
	Arrays.fill(prime, 1);
	prime[0]=0;
	prime[1]=0;
	for(int i=2;i*i<=1000005;i++) {
		if(prime[i]==1)
		for(int j=i*i;j<=1000005;j+=i) {
			prime[j]=0;
		}
	}
	for(int i=1;i<1000005;i++) {prime[i]=prime[i]+prime[i-1];}
}
static void sort(long [] a) {
  ArrayList<Long> aa = new ArrayList<>();
  for (long  i : a) {
    aa.add(i);
  }Collections.sort(aa);  for (int i = 0; i < a.length; i++)
    a[i] = aa.get(i);	  }
static void sort(int [] a) {
  ArrayList<Integer> aa = new ArrayList<>();
  for (int  i : a) {
    aa.add(i);
  } Collections.sort(aa); for (int i = 0; i < a.length; i++)
    a[i] = aa.get(i); }
static long pow(long x, long y) {
  long res = 1l;
  while (y != 0) {
    if (y % 2 == 1) {
      res = (x%mod * res%mod)%mod; } y /= 2; x = (x%mod * x%mod)%mod;
  } return res%mod;  }

//>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Fast IO <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< //

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;
  }}
}
class two implements Comparable<two>{
	int val;int ind;
	two(int val,int ind){
		this.val=val;
		this.ind=ind;
	}
	@Override
	public int hashCode() {
		final int prime = 31;
		int result = 1;
		result = prime * result + ind;
		result = prime * result + val;
		return result;
	}
	@Override
	public boolean equals(Object obj) {
		if (this == obj)
			return true;
		if (obj == null)
			return false;
		if (getClass() != obj.getClass())
			return false;
		two other = (two) obj;
		if (val != other.val)
			return false;
		return true;
	}
	@Override
	public int compareTo(two o) {
	
		return -Integer.compare(o.val, this.val);
	}

}


/*
 * imp log method using math function is somewhat inaccurate to check power of
 * two
 *
 */