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 <vector>
#define sz(x) (int)x.size()
using namespace std;

int32_t main() {
    int q;
    cin >> q;
    while (q--) {
        int n;
        cin >> n;
        string a, b;
        cin >> a >> b;
        vector<int> av, bv;
        int cnt0 = 0;
        int cnt1 = 0;
        for (int i = 0; i < n; ++i) {
            if (a[i] == '1') {
                ++cnt1;
            }
            else {
                ++cnt0;
            }
            av.push_back(a[i] - '0');
            bv.push_back(b[i] - '0');
        }
        bool isrev = 0;
        bool ans = 1;
        for (int i = n - 1; i >= 0; --i) {
            int old = av[i];
            if (isrev) {
                av[i] = !av[i];
            }
            if (av[i] != bv[i]) {
                if (cnt1 == cnt0) {
                    isrev = !isrev;
                }
                else {
                    ans = 0;
                }
            }
            if (old == 1) {
                --cnt1;
            }
            else {
                --cnt0;
            }
        }
        if (ans) {
            cout << "YES\n";
        }
        else {
            cout << "NO\n";
        }
       
    }
}