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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define N 200005
using namespace std;
typedef long long ll;

ll n, ans, a[N], d[N][3];
int main() {
	int i;
	memset(d, 0xc0, sizeof(d));
	cin >> n;
	for(i=1; i<=n; i++) scanf("%lld", &a[i]);
	d[1][0] = a[1];
	d[2][1] = a[2];
	d[3][0] = a[1] + a[3];
	d[3][2] = a[3];
	for(i=4; i<=n; i++) {
		d[i][0] = d[i-2][0] + a[i];
		d[i][1] = max(d[i-3][0], d[i-2][1]) + a[i];
		d[i][2] = max({d[i-4][0], d[i-3][1], d[i-2][2]}) + a[i];
	}
	if(n&1) cout << max({d[n-2][0], d[n-1][1], d[n][2]});
	else cout << max(d[n-1][0], d[n][1]);
	return 0;
}