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.
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#include<queue>
#include<vector>
#include<map>
#define oo 0x3f3f3f3f
#define OO 0x3f3f3f3f3f3f3f3f
#define LL long long 
#define res register int
#define STP system("pause") 
using namespace std;
const int N=3e5;
int One[N+5],Zro[N+5],c[N+5];
int main(){
	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		string a,b;
//		int c[N+]
		cin>>a>>b;
		for(int i=1;i<=n;++i){
			One[i]=One[i-1]+(a[i-1]=='1'?1:0);
			Zro[i]=Zro[i-1]+(a[i-1]=='0'?1:0);
			if(a[i-1]==b[i-1])	c[i]=0;
			else c[i]=1;
		}
//		for(int i=1;i<=n;++i)	cout<<c[i]<<' ';
//		cout<<'\n';
//		for(int i=1;i<=n;++i)	cout<<One[i]<<' ';
//		cout<<'\n';
//		for(int i=1;i<=n;++i)	cout<<Zro[i]<<' ';
//		cout<<'\n';
		int flg=0,lst=0;
		c[n+1]=3;
		for(int i=1;i<=n+1;++i){
			if(i==1||c[i]==c[i-1])	continue;
			else{
				if(One[i-1]-One[lst]==Zro[i-1]-Zro[lst]){
					lst=i-1;
					continue;
				}
				else if(i==n+1&&c[i-1]==0)	continue;
				else{
//					cout<<i<<'*'<<'\n';
					flg=1;
					break;
				}
			}
		}
		if(flg)	cout<<"NO\n";
		else cout<<"YES\n";
	}
	return 0;
}