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;
typedef     long long    ll;
typedef     vector<int> vi;
typedef     vector<long long> vl;
typedef     pair<int, int>pi;
typedef     pair<long long, long long>pl;
#define F   first
#define S   second
#define pb  push_back
#define     all(x)      x.begin() , x.end()
#define mp  make_pair
#define     REP(i,a,b) for(i=a;i<=b;i++)
#define     mem(a)      memset(a , 0 ,sizeof a)
#define     memn(a)     memset(a , -1 ,sizeof a)
int main()
{
    int test = 1;
     cin >> test;
    while (test--)
    {
        
         int i,j,n,f2=1,O=0,Z=0;
         cin>>n;
          string s1,s2;
         cin>>s1>>s2;
         int o[n+2],z[n+3];
         for(i=0;i<n;i++)
         {
             if(s1[i]=='1') O++;
             else Z++;
             o[i]=O;
             z[i]=Z;
         }
         int f=1;
         for(i=n-1;i>=0;)
         {
             if(s1[i]==s2[i])
             {
                   if(f==0&&o[i]!=z[i])
                    {
                        f2=0; break;
                    }
                    while(i>=0&&s1[i]==s2[i]) i--;
                    if(s1[i]!=s2[i]) f=1;
             }else
             {
                if(f==1&&o[i]!=z[i])
                    {
                        f2=0; break;
                    }
                    while(i>=0&&s1[i]!=s2[i]) i--;
                    if(s1[i]==s2[i]) f=0;
             }
         }
      if(f2==0) cout<<"NO"<<endl  ;
      else cout<<"YES"<<endl;
    }

    return 0;
}