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 read_vc(vector<int> &vc, int n)
{
        for (int i = 0; i < n; ++i)
                cin >> vc[i];
}

void solve(){
        string s1, s2;
        int n;
        cin >> n;
        cin >> s1 >> s2;
        int first = 0, second = 0;
        for(char c : s1){
                if(c == '1')first++;
        }
        for(char c : s2){
                if(c == '1')second++;
        }
        if(first != second){
                cout<<"NO\n";
                return;
        }
        
        int count = 0, zero = n - first;

        for(int i = n - 1; i >= 0; --i){
                char ch = s1[i];
                if(count % 2 != 0){
                        if(ch == '1')ch = '0';
                        else ch = '1';
                }
                
                if(ch != s2[i]){
                        //if(i == 3)cout<<zero <<' '<<first<<endl;
                        if(zero != first){
                                cout<<"NO\n";
                                //cout<<zero<<" "<<first<<" "<<i<<endl;
                                return;
                        }
                        count++;
                        if(ch == '1')ch = '0';
                        else ch = '1';
                }
                        //cout<<ch<<endl;
                
                        if(ch == '1')first--;
                        else zero--;
        }
        cout<<"YES\n";
}



int main()
{
        //test summary https://app.codility.com/c/feedback/FRFDN6-8C4/
        //freopen("candies.in","r",stdin);
        //freopen("candies.out","w",stdout);
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        int tc = 1;
        cin >> tc;

        while (tc--)
        {
                solve();
        }
        
        return 0;
}