#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> ii;

void solve(){
	
	string a, b;
	int n;
	cin >> n;
	cin >> a >> b;
	vector<int> soma(n, 0);
	
	for(int i = 0; i < n; i++){
		soma[i] += a[i] == '1';
		if(i)
			soma[i] += soma[i-1];
	}
	
	bool inv = false, flag = true;
	for(int i = n-1; i >= 0; i--){
		if((!inv && a[i] != b[i]) || (inv && a[i] == b[i])){
			if(soma[i] * 2 == i+1)
				inv = !inv;
			else{
				flag = false;
				//cout << i << endl;
			}
 		}
	}
	
	cout << (flag ? "YES" : "NO") << endl;
}

int32_t main(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	
	//freopen("in", "r", stdin);
	//freopen("out", "w", stdout);
	
	int tc = 1;
	cin >> tc;
	while(tc--)
		solve();
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> ii;

void solve(){
	
	string a, b;
	int n;
	cin >> n;
	cin >> a >> b;
	vector<int> soma(n, 0);
	
	for(int i = 0; i < n; i++){
		soma[i] += a[i] == '1';
		if(i)
			soma[i] += soma[i-1];
	}
	
	bool inv = false, flag = true;
	for(int i = n-1; i >= 0; i--){
		if((!inv && a[i] != b[i]) || (inv && a[i] == b[i])){
			if(soma[i] * 2 == i+1)
				inv = !inv;
			else{
				flag = false;
				//cout << i << endl;
			}
 		}
	}
	
	cout << (flag ? "YES" : "NO") << endl;
}

int32_t main(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	
	//freopen("in", "r", stdin);
	//freopen("out", "w", stdout);
	
	int tc = 1;
	cin >> tc;
	while(tc--)
		solve();
}

```