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>
#include <string>
#include <map>
#include <memory>
#include <algorithm>

using namespace std;

int main() {
	int T;
	cin >> T;
	for (int t = 0; t < T; t++) {
		int N;
		cin >> N;
		string s1, s2;
		cin >> s1 >> s2;
		vector<int> diff(s1.size(), 0);
		if (s1[0] == '1')
			diff[0] = 1;
		else
			diff[0] = -1;
		for (int i = 1; i < s1.size(); i++) {
			if (s1[i] == '1')
				diff[i] = 1;
			else
				diff[i] = -1;
			diff[i] += diff[i - 1];
		}
		int x = 0;
		bool ok = true;
		for (int i = s1.size() - 1; i >= 0; i--) {
			int a = s1[i] - '0';
			int b = s2[i] - '0';
			if ((a + x + b) % 2 != 0) {
				if (diff[i] != 0) {
					ok = false;
					break;
				}
				else {
					x = (x + 1) % 2;
				}
			}
		}
		cout << (ok ? "YES\n" : "NO\n");
	}
	return 0;
}