Given is an integer sequence A_1, ..., A_N of length N.

We will choose exactly \left\lfloor \frac{N}{2} \right\rfloor elements from this sequence so that no two adjacent elements are chosen.

Find the maximum possible sum of the chosen elements.

Here \lfloor x \rfloor denotes the greatest integer not greater than x.

Constraints

* 2 \leq N \leq 2\times 10^5
* |A_i|\leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 ... A_N


Output

Print the maximum possible sum of the chosen elements.

Examples

Input

6
1 2 3 4 5 6


Output

12


Input

5
-1000 -100 -10 0 10


Output

0


Input

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000


Output

5000000000


Input

27
18 -28 18 28 -45 90 -45 23 -53 60 28 -74 -71 35 -26 -62 49 -77 57 24 -70 -93 69 -99 59 57 -49


Output

295
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int N = 200005;
const ll inf = 1e17;
int n, a[N];
ll d[N], r[N], e[N], ma;

ll E(int n) {
	return e[n];
}

ll D(int n) {
	if(n == 1) return 0;
	else if(n & 1) return max(D(n - 2) + a[n], E(n - 1));
	else return E(n);
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	for(int i = 1; i <= n; ++i)
		cin >> a[i];
	for(int i = 2; i <= n; i += 2) {
		d[i] = d[i - 2] + a[i - 1];
		r[i] = r[i - 2] + a[i];
	}
	ma = 0;
	for(int i = 2; i <= n; i += 2) {
		ma = max(ma, d[i] - r[i]);
		e[i] = ma + r[i];
	}
	cout << D(n) << endl;
	return 0;
}
