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<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<bitset>
#include<queue>
#include<stack>
#include<map>
using namespace std;

const long long maxn=300020,maxm=11000,inf=1e18;
const long long mod=1000000007;

typedef long long ll;
int t;         
int tag[300020],a[300020],b[300020];
char sa[300020],sb[300020];


int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,cnt=0,gt=0;
		scanf("%d",&n);
		scanf("%s",sa+1);
		for(int i=1;i<=n;i++) 
		{
			a[i]=sa[i]-'0';
			cnt+=(sa[i]-'0');
			if(i%2==0&&cnt==i/2) tag[i]=1;
			else tag[i]=0;
		}
		scanf("%s",sb+1);
		for(int i=1;i<=n;i++)
		{
			b[i]=sb[i]-'0';
		}
		int now=0,flag=1;
		for(int i=n;i>=1;i--)
		{
			if(tag[i])
			{
				if((a[i]^now)!=b[i]) now^=1;
			}
			else
			{
				if((a[i]^now)!=b[i])
				{
					flag=0;
					break;
				}
			}
		}
		if(flag) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}