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>
#include<iomanip>
#include<string>
#include<vector>
#include<algorithm>
#include<assert.h>
#include<cmath>
#include<unordered_set>

using namespace std;


void solve() {
    int n;
    cin >> n;
    //vector<bool> str;
    //for (size_t i = 0; i < n; ++i) {
    //    char c;
    //    cin >> c;
    //    str[i] = c - '0';
    //}
    //vector<bool> rev = str;
    //rev.flip();
    string a;
    string b;
    cin >> a >> b;

    size_t count_0 = 0;
    count_0 += a[0] == '0';
    for (size_t i = 1; i < n; ++i) {
        if ((a[i - 1] == b[i - 1]) ^ (a[i] == b[i])) {
            if (count_0 * 2 != i) {
                cout << "NO\n";
                return;
            }
        }
        count_0 += a[i] == '0';
    }
    if (a.back() != b.back()) {
        if (count_0 * 2 != a.size()) {
            cout << "NO\n";
            return;
        }
    }
    cout << "YES\n";
    return;
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    cin >> t;
    
    while (t--) solve();

    return 0;
}

/*
1
2 7 25
*/
