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 <cmath>
#include <algorithm>
#include <iomanip>
#include <set>
#include <map>
#include <queue>
typedef long long ll;

using namespace std;



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

	int t;
	cin >> t;
	for (int z = 0; z < t; ++z) {
		int n;
		cin >> n;
		string s1, s2;
		cin >> s1 >> s2;
		if (n == 1 && s1 != s2) {
			cout << "NO\n";
			continue;
		}
		vector<pair<int, int>> p1(n + 1), p2(n + 1);
		for (int i = 0; i < n; ++i) {
			p1[i + 1] = p1[i];
			p2[i + 1] = p2[i];
			if (s1[i] == '1') {
				p1[i + 1].first++;
			}
			else {
				p1[i + 1].second++;
			}
			if (s2[i] == '1') {
				p2[i + 1].first++;
			}
			else {
				p2[i + 1].second++;
			}
		}
		vector<int> men(n), raz(n + 2);
		bool no = 0;
		for (int i = 1; i <= n; ++i) {
			if (p1[i].first == p1[i].second) {
				men[i - 1] = 1;
			}
			if (!((p1[i].first == p2[i].second && p1[i].second == p2[i].first) || 
				(p1[i].second == p2[i].second && p1[i].first == p2[i].first))) {
				no = 1;
				break;
			}
			if (p1[i].first != p2[i].first) {
				raz[i - 1] = 1;
			}
		}
		if (no) {
			cout << "NO\n";
			continue;
		}
		bool mef = 1;
		for (int i = 0; i < n; ++i) {
			if (men[i] == 1) {
				mef = 0;
				break;
			}
		}
		if (mef && s1 != s2) {
			cout << "NO\n";
			continue;
		}
		int rm = -1;
		for (int i = 0; i < n; ++i) {
			if (men[i]) {
				rm = i;
			}
		}
		for (int i = rm + 1; i < n; ++i) {
			if (raz[i]) {
				no = 1;
				break;
			}
		}
		if (no) {
			cout << "NO\n";
			continue;
		}
		int last = raz[0];
		for (int i = 0; i < n; ++i) {
			if (men[i]) {
				last = raz[i + 1];
				continue;
			}
			if (raz[i] != last) {
				no = 1;
				break;
			}
			else {
				last = raz[i];
			}
		}
		if (no) {
			cout << "NO\n";
			continue;
		}
		else {
			cout << "YES\n";
		}
	}
	return 0;
}