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 <iostream>
using namespace std;
#include<unordered_map>
#include<map>
#include<string>
#include<vector>
#include<algorithm>
#include<math.h>
#include<climits>
#include<sstream>
#include<queue>
#include<stack>
#include<set>
#define ll int
#define loop(k,x,n) for( long long x = k; x < n; ++x )
#define rloop(k,x,n) for( long long x = n-1-k ; x>=0 ; x--)
#define t(n) long long* t = new long long[n]
#define h(n) long long* h = new long long[n]
#define visit(n) long long* visit = new long long[n]
#define v(x) vector<x> v
#define um(x,y) unordered_map<x,y> um;
#define endl '\n'


void solve(){

   ll n;
   cin>>n;


   string a,b;
   cin>>a>>b;

   vector<pair<ll,ll>> v;

   ll zero = 0;
   ll one = 0;

   for( ll i=0 ; i<n ; i++ ){
    if( a.substr(i,1) == "0" ){
        zero++;
    }else{
        one++;
    }
    v.push_back({zero,one});
   }

   bool done =true;
   ll x = 0;

   for( ll i=n-1 ; i>=0 ; i-- ){
       string h = a.substr(i,1);
       if( x%2 !=0 ){
           if( h == "0" ){
               h ="1";
           }else{
               h = "0";
           }
       }
       if( h == b.substr(i,1) ){
           continue;
       }else{
         x++;
         if( v[i].first == v[i].second ){
             continue;
         }else{
             done = false;
             break;
         }
       }

   }

   if( done ){
       cout<<"YES"<<endl;
   }else{
       cout<<"NO"<<endl;
   }







}

int main(){

  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  ll t;
  cin>>t;

  while( t > 0 ){
   solve();
   t--;
  }

}
