#include <bits/stdc++.h>
#define ll long long int
#define l long int
#define ld long double
using namespace std;
const ll mod = 1e9 + 7;
int c=0;

int main()
{
   ll t,n,i,sum;
   bool ans;
   string a,b;
   cin>>t;
   while(t--)
   {
    cin>>n>>a>>b;
     vector<int>h(n,0);
     sum=0;
     ans=true;
     for(i=0;i<n;i++)
     {
        sum+=a[i]-'0';
        if(i+1 == 2*sum) h[i]=1;
     }
     if(a[n-1] != b[n-1])
     {
        if(h[n-1]==0) 
         ans =false;
     }
     for(i=0;i<n-1;i++)
     {
         if((a[i] != b[i] && a[i+1] ==b[i+1]) || (a[i]==b[i] && a[i+1] != b[i+1]))
         {
            if(h[i]==0) ans = false;
         }
         
     }
      if(ans==false) cout<<"NO"<<endl;
      else cout<<"YES"<<endl;

   }
}
### Prompt

Your challenge is to write a CPP solution 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>
#define ll long long int
#define l long int
#define ld long double
using namespace std;
const ll mod = 1e9 + 7;
int c=0;

int main()
{
   ll t,n,i,sum;
   bool ans;
   string a,b;
   cin>>t;
   while(t--)
   {
    cin>>n>>a>>b;
     vector<int>h(n,0);
     sum=0;
     ans=true;
     for(i=0;i<n;i++)
     {
        sum+=a[i]-'0';
        if(i+1 == 2*sum) h[i]=1;
     }
     if(a[n-1] != b[n-1])
     {
        if(h[n-1]==0) 
         ans =false;
     }
     for(i=0;i<n-1;i++)
     {
         if((a[i] != b[i] && a[i+1] ==b[i+1]) || (a[i]==b[i] && a[i+1] != b[i+1]))
         {
            if(h[i]==0) ans = false;
         }
         
     }
      if(ans==false) cout<<"NO"<<endl;
      else cout<<"YES"<<endl;

   }
}
```