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;
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int ntest;
    cin >> ntest;
    for(int z = 1;z <= ntest; z++)
    {
        int n;
        string a, b;
        cin >> n >> a >> b;
        bool check = true;
        int wrong = 0;
        int dem0 = 0;
        int dem1 = 0;
        int len = 0;
        for(int i = 0; i < n; i++)
            {
                len++;
                if(a[i] != b[i]) wrong++;
                if(a[i] == '0') dem0++;
                else dem1++;
                //cout << i << " " << dem0 << " " << dem1 << endl;
                if(dem0 == dem1)
                {
                    dem0 = dem1 = 0;
                    if(wrong == len)
                    {
                        //cout << 4 << endl;
                        for(int j = i - len + 1; j <= i; j++)
                            {
                                int t = a[j];
                                t -= 48;
                                t = 1 - t;
                                a[j] = (t + 48);
                            }
                        //cout << i << " " << len << " " << wrong << endl;
                        //cout << a << endl << b << endl;
                        wrong = 0;
                        len = 0;
                        continue;
                    }
                    if(wrong == 0)
                    {
                        len = 0;
                        continue;
                    }
                }
            }
        //cout << a << endl << b << endl;
        if(a != b) cout << "NO" << '\n';
        else cout << "YES" << '\n';
    }
    return 0;
}
