#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define M 1000000007

signed main(){
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

	int t;
	cin >> t;
	while(t--){
		int n;
		string s1,s2;
		cin >> n >> s1 >> s2;

		int a=0, b=0;
		for(int i=0; i<n; i++){
			if(s1[i]=='0')
				a++;
			else
				b++;
		}

		int f=0;
		bool ans=1;
		for(int i=n-1; i>=0; i--){
			if(f==0){
				if(s1[i]==s2[i]){
					if(s1[i]=='0')
						a--;
					else
						b--;
					continue;
				}

				if(a!=b){
					ans=0;
					break;
				}
				else{
					f=1;
					// swap(a,b);
					if(s1[i]=='0')
						a--;
					else
						b--;
				}
			}
			else{
				if(s1[i]!=s2[i]){
					if(s1[i]=='0')
						a--;
					else
						b--;
					continue;
				}

				if(a!=b){
					ans=0;
					break;
				}
				else{
					f=0;
					// swap(a,b);
					if(s1[i]=='0')
						a--;
					else
						b--;
				}
			}
		}

		if(ans)
			cout << "YES\n";
		else
			cout << "NO\n";
	}   
    

    return 0;
}
### Prompt

Please create a solution in cpp to the following problem:

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;
#define int long long
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define M 1000000007

signed main(){
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

	int t;
	cin >> t;
	while(t--){
		int n;
		string s1,s2;
		cin >> n >> s1 >> s2;

		int a=0, b=0;
		for(int i=0; i<n; i++){
			if(s1[i]=='0')
				a++;
			else
				b++;
		}

		int f=0;
		bool ans=1;
		for(int i=n-1; i>=0; i--){
			if(f==0){
				if(s1[i]==s2[i]){
					if(s1[i]=='0')
						a--;
					else
						b--;
					continue;
				}

				if(a!=b){
					ans=0;
					break;
				}
				else{
					f=1;
					// swap(a,b);
					if(s1[i]=='0')
						a--;
					else
						b--;
				}
			}
			else{
				if(s1[i]!=s2[i]){
					if(s1[i]=='0')
						a--;
					else
						b--;
					continue;
				}

				if(a!=b){
					ans=0;
					break;
				}
				else{
					f=0;
					// swap(a,b);
					if(s1[i]=='0')
						a--;
					else
						b--;
				}
			}
		}

		if(ans)
			cout << "YES\n";
		else
			cout << "NO\n";
	}   
    

    return 0;
}
```