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;

void solve(string a, string b, int n) {
    int i = 0;
    //cout << "a: " << a << endl;
    //cout << "b: " << b << endl;
    while(i < n) {
        int zcnt = 0, ocnt = 0;
        int j = 0;
        while(i + j < n && ((zcnt == 0 && ocnt == 0) || zcnt != ocnt)) {
            zcnt += 1 - (a[i + j] - '0');
            ocnt += a[i + j] - '0';
            //cout << "z: " << zcnt << " o: " << ocnt << endl;
            j++;
        }
        //cout << "eqi: " << i + j << endl;
        //if(i + j < n)
        //    j++;
        bool ok = false;
        if(a.substr(i, j)  == b.substr(i, j)) {
            ok = true;
        }

        if(!ok) {
            if(i + j == n && zcnt != ocnt) {
                cout << "NO" << endl;
                return;
            }
            for(int k = 0; k < j; k++) {
                if(1 - (a[i + k] - '0') + '0' != b[i + k]) {
                    cout << "NO" << endl;
                    return;
                }
            }
        }
        i += j;
    }
    cout << "YES" << endl;
}


int main() {
    int T;
    cin >> T;
    while(T--){
        int n;
        string a, b;
        cin >> n >> a >> b;
        solve(a, b, n);
    }
}
    
