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.
/* In the name of *Allah* *
 *   Thank You *Parsa*    */
#include<iostream>
#include<algorithm>
#include<math.h>
#include<string>
#include<ios>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<numeric>
using namespace std;

#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()

typedef vector <int> VI;
typedef vector <long long> VLL;
typedef pair <int, int> PII;
typedef set <int> SI;
typedef set <long long> SLL;
typedef long long ll;
typedef long double ld;

const int inf_int = 1 << 30;
const ll inf_longlong = 1LL << 62;
const int MOD = 1e9 + 7;

const int N = 3e5 + 5;
int n;
string a, b;

void read_input() {
	cin >> n >> a >> b;
}

void solve() {
	int cnt1 = 0, cnt2 = 0;
	VI V;
	V.pb(-1);
	for (int i = 0; i < n; i++) {
		cnt1 += (a[i] == '0'), cnt2 += (a[i] == '1');
		if (cnt1 == cnt2) V.pb(i);
	}

	int k = V.back();
	for (int i = n - 1; i > k; i--) {
		if (a[i] != b[i]) {
			cout << "No" << '\n';
			return;
		}
	}

	reverse(all(V));
	for (int i = 0; i < sz(V) - 1; i++) {
		int cond1 = 0, cond2 = 0;
		for (int j = V[i]; j > V[i + 1]; j--) {
			if (a[j] != b[j]) cond1++;
			else cond2++;
		}
		if (cond1 && cond2) {
			cout << "No" << '\n';
			return;
		}
	}
	cout << "Yes" << '\n';
}

void write_output() {
}

int main() {
	ios:: sync_with_stdio(false), cin.tie(0), cout.tie(0);
#ifndef ONLINE_JUDGE
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
#endif

	int test_cases = 1; cin >> test_cases;
	while (test_cases--) {
		read_input(), solve(), write_output();
	}
	return 0;
}
// Thanks *Allah*