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;
#define ll long long int
#define endl "\n"
#define mod 1000000007
#define maxN 100000





int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
// #ifndef ONLINE_JUDGE
// 	{
// 		freopen("input.txt", "r", stdin);
// 		freopen("output.txt", "w", stdout);
// 	}
// #endif
    ll t;
    cin>>t;
    while(t--){
        ll n;
        cin>>n;
        string a,b;
        cin>>a>>b;
        if(a==b){
            cout<< "YES"<< endl;
            continue;
        }
        ll one=0,zero=0;
        set<ll> s;
        for(ll i=0;i<n;i++){
            if(a[i]=='0'){
                zero++;
            }else{
                one++;
            }
            if(one==zero){
                s.insert(i);
            }
        }
        if(s.size()==0){
            cout<< "NO"<<endl;
        }else{
            bool flag=true,f_1=true;
            for(ll i=n-1;i>=0;i--){
                if(flag){
                    if(a[i]!=b[i]){
                        if(s.find(i)==s.end()){
                            f_1=false;
                            break;
                        }
                        flag=false;
                    }
                }else{
                    if(a[i]==b[i]){
                        if(s.find(i)==s.end()){
                            f_1=false;
                            break;
                        }
                        flag=true;
                    }
                }
            }
            if(f_1){
                cout<< "YES"<<endl;
            }else{
                cout<< "NO"<<endl;
            }
        }
    }
	return 0;

}