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;

void solve()
{
        int sz;
        cin>>sz;
        
        string a,b;
        cin>>a>>b;
        
        
        if(a==b)
        {
            cout<<"YES";
            return;
        }
        
        int cnt=0;
        
        vector<int> vec;
        
        vec.push_back(-1);
        
        for(int i=0;i<sz;i++)
        {
            if(a[i]=='1')
            cnt++;
            else
            cnt--;
            
           
            if(cnt==0)
            {
                vec.push_back(i);
            }
            
        }
        
        bool flag=1;
        
        sort(vec.begin(),vec.end());  
       
        if(vec.size()==1)
        {
            flag=(a==b);
        }
        else if(vec.size()>1)
        {
            int n=vec.size();
            
            if(vec[n-1]!=sz-1)
            {
                string act = a.substr(vec[n-1]+1);
                string tocomp = b.substr(vec[n-1]+1);
                
                string convt;
                
                
                
                if(act!=tocomp )
                {
                    flag=0;
                }
            }
            
            
            for(int i=n-1;i>0;i--)
            {
                int len=vec[i]-vec[i-1];
                
                string act = a.substr(vec[i-1]+1,len);
                string tocomp = b.substr(vec[i-1]+1,len);
                
                string convt;
                
                for(int k=vec[i-1]+1;k<=vec[i];k++)
                {
                    if(a[k]=='1')
                    convt.push_back('0');
                    else
                    convt.push_back('1');
                }
                
                
                if(act!=tocomp && tocomp!=convt)
                {
                    flag=0;
                    break;
                }
            }   
            
            
        }
        
        
        
        (flag?cout<<"YES":cout<<"NO");
        
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        solve();
        cout<<endl;
    }
    
    return 0;
}



