#include<bits/stdc++.h>
using namespace std ;
int main()
{
      int t;
      cin>>t;
      while(t--)
      {
            int n,a,b=0,i,j,one=0,zero=0;
            cin>>n;
            string s1,s2;
            cin>>s1>>s2;
            vector<pair<int,int>>v;
            j=0;
            
            for(i=0;i<n;i++)
            {
                  if(s1[i]=='0')
                  zero++;
                  if(s1[i]=='1')
                  one++;
                  
                  if(zero==one)
                  {
                        v.push_back(make_pair(j,i));
                        j=i+1;
                  }
                  
            }
            for(auto x : v)
            {
                  i=x.first;
                  j=x.second;
                  if(s1[i]==s2[i])
                  continue;
                  else
                  {
                        for(a=i;a<=j;a++)
                        {
                              if(s1[a]=='0')
                              s1[a]='1';
                              else
                              {
                                    s1[a]='0';
                              }
                        }
                  }
            }
            // for(a=0;a<n;a++)
            // {
            //       if(s1[a]!=s2[a])
            //       {
            //             b=1;
            //             break;
            //       }
            // }
            if(s1!=s2)
            cout<<"NO"<<endl;
            else
            cout<<"YES"<<endl;
      }
}
### Prompt

Generate 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>
using namespace std ;
int main()
{
      int t;
      cin>>t;
      while(t--)
      {
            int n,a,b=0,i,j,one=0,zero=0;
            cin>>n;
            string s1,s2;
            cin>>s1>>s2;
            vector<pair<int,int>>v;
            j=0;
            
            for(i=0;i<n;i++)
            {
                  if(s1[i]=='0')
                  zero++;
                  if(s1[i]=='1')
                  one++;
                  
                  if(zero==one)
                  {
                        v.push_back(make_pair(j,i));
                        j=i+1;
                  }
                  
            }
            for(auto x : v)
            {
                  i=x.first;
                  j=x.second;
                  if(s1[i]==s2[i])
                  continue;
                  else
                  {
                        for(a=i;a<=j;a++)
                        {
                              if(s1[a]=='0')
                              s1[a]='1';
                              else
                              {
                                    s1[a]='0';
                              }
                        }
                  }
            }
            // for(a=0;a<n;a++)
            // {
            //       if(s1[a]!=s2[a])
            //       {
            //             b=1;
            //             break;
            //       }
            // }
            if(s1!=s2)
            cout<<"NO"<<endl;
            else
            cout<<"YES"<<endl;
      }
}
```