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>
#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <cstring>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#include <list>
#include <string>
#include <vector>
#include <algorithm>
#include <sstream>
#include <unordered_map>
using namespace std;
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pll pair<ll,ll>
#define pii pair<int,int>
#define bg begin
#define rbg rbegin
#define ed end
#define endl '\n'
#define dbg(x) cout << #x << "===" << x << endl
typedef double db;
typedef long long ll;
typedef unsigned long long ull;

const int N = 3e5;
int a[N], b[N], vis[N];

int gcd(int a, int b) {
	return !b ? a : gcd(b, a % b);
}

ll qpow(ll a, ll b, ll mod) {
	ll ans = 1;
	while (b) {
		if (b & 1)	ans = (ans * a) % mod;
		a = (a * a) % mod, b >>= 1;
	}
	return ans % mod;
}

int main() {

	//freopen("D:\\测试用\\in.txt", "r", stdin);
	//freopen("D:\\测试用\\out.txt", "w+", stdout);
	int t; cin >> t;
	while (t--) {
		int n; cin >> n;
		int sum0, sum1;
		sum0 = sum1 = 0;
		rep(i, 1, n) {
			scanf("%1d", &a[i]);
			vis[i] = 0;
			if (a[i])	sum1++;
			else sum0++;
		}
		rep(i, 1, n)	scanf("%1d", &b[i]);
		per(i, n, 1) {
			if (sum0 == sum1) vis[i] = 1;
			if (a[i])	sum1--;
			else sum0--;
		}
		int f = 1;
		int sum = 0;
		per(i, n, 1) {
			if (sum & 1)	a[i] ^= 1;
			if (a[i] != b[i]) {
				if (vis[i]) {
					a[i] ^= 1;
					sum++;
				}
				else f = 0;
			}
		}
		if (f)	cout << "YES" << endl;
		else cout << "NO" << endl;
	}
	return 0;
}