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;
const int MAXN = 4e5 + 5;
#define ll long long
int n;
ll d[MAXN], f[MAXN][2];
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%lld", d + i);
	f[1][1] = -1e18;
	for (int i = 2; i <= n; i ++) {
		if (i & 1) {
			f[i][1] = max(max(f[i - 3][1], f[i - 3][0]), max(f[i - 2][1], f[i - 2][0])) + d[i];
			f[i][0] = f[i - 2][0] + d[i - 1];
		}
		else {
			f[i][1] = max(f[i - 2][1], f[i - 2][0]) + d[i];
			f[i][0] = f[i - 2][0] + d[i - 1];
		} 
	} 
	printf("%lld\n", max(f[n][0], f[n][1]));
	return 0;
} 