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<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define YES cout<<"YES"
#define NO cout<<"NO"
#define Yes cout<<"Yes"
#define No cout<<"No"
#define yes cout<<"yes"
#define no cout<<"no"
#define FOR(S , E) for(ll i=S;i<=E;i++)
#define ROF(E , S) for(ll i=E;i>=S;i--)
// V.push_back() // V.pop_front()// V.top()
//V.pop()//V.front() // V.back() // pop_back()
//second // first // insert // continue // break
// erase // St.push() // V.size() // S.length()
// clear() // S.length() // reverse // greater // push

int main()
{
 
 

#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	
	int T;
	cin>>T;
	while(T--)
	{
		int N;
		cin>>N;
		string S1 , S2;
		cin>>S1>>S2;
		map<char,char>Map;
		Map['0'] = '0';
		Map['1'] = '1';
		int Zeros = 0 , Ones = 0;
		for(char C : S1)
		{
			if(C == '1')Ones++;
			else Zeros++;
		}
		bool Cant = 0;
		for(int i=N-1;i>=0;i--)
		{
			if(Map[S1[i]] == S2[i])
			{
				if(Map[S1[i]] == '1')Ones--;
				else Zeros--;
				continue;
			}

			if(Map[S1[i]]!=S2[i] && Zeros == Ones)
			{
				Map['0'] = Map['0'] == '0'?'1':'0';
				Map['1'] = Map['1'] == '1'?'0':'1';

				if(Map[S1[i]] == '1')Ones--;
				else Zeros--;
			}
			else if(Map[S1[i]]!=S2[i] && Zeros != Ones)
			{
				Cant = 1;
				break;
			}
		}
		if(Cant)NO;
		else YES;
		cout<<endl;
	}



	return 0;
} 
