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;

#define int 	       long long
#define endl	       "\n"
#define vi             vector <int>
#define pb             push_back
#define all(x)         (x).begin(),(x).end()
#define fill(a,b)      memset(a,b,sizeof(a))
const int mod = 1e9 + 7;

const int N = 1e5 + 5;

void solve() {
	int n;
	cin >> n;
	string a, b;
	cin >> a >> b;
	int ac0 = 0, ac1 = 0, bc0 = 0, bc1 = 0;
	vi v;
	v.pb(-1);

	for (int i = 0; i < n; i++) {
		if (a[i] == '0')
			ac0++;
		else
			ac1++;

		if (ac0 == ac1) {
			v.pb(i);
		}
	}

	// for (int i = 0; i < n; i++) {
	// 	if (b[i] == '0')
	// 		bc0++;
	// 	else
	// 		bc1++;
	// }
	// for (auto it : v) {
	// 	cout << it << " ";
	// }
	// cout << endl;

	// if (ac1 != bc1 || ac0 != bc0) {
	// 	cout << "NO" << endl;
	// 	return;
	// }
	int p = 0;
	vi v1;
	for (int i = v.size() - 1; i >= 1; i--) {
		char c;
		c = a[v[i]];
		if (p % 2 == 1) {
			if (a[v[i]] == '1') {
				c = '0';
			}
			else {
				c = '1';
			}
		}
		if (c != b[v[i]]) {
			p++;
		}
		v1.pb(p);
	}

	reverse(all(v1));

	int k = v1.size() - 1;
	for (int i = v.size() - 1; i >= 1; i--) {
		int j = v[i];
		if (v1[k] % 2 == 1) {
			while (j > v[i - 1]) {
				if (a[j] == '1') {
					a[j] = '0';
				}
				else {
					a[j] = '1';
				}
				j--;
			}
		}
		k--;
	}

	if (a == b) {
		cout << "YES" << endl;
	}
	else
		cout << "NO" << endl;

}

int32_t main()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	int test_case = 1;
	cin >> test_case;
	while (test_case--) {
		solve();
	}
	return 0;
}