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 <stdio.h>
#include <algorithm>
#include <climits>

long long input[200010];
long long dp[3][200010];
bool check[3][200010];
int n;

long long func(int left, int x) {
	if (x > n) {
		if (left <= 1)
			return 0;
		return LLONG_MIN / 2;
	}
	if (check[left][x] == true)
		return dp[left][x];

	check[left][x] = true;
	dp[left][x] = func(left, x + 2) + input[x];

	if (left > 0) {
		dp[left][x] = std::max(dp[left][x], func(left - 1, x + 1));
	}

	return dp[left][x];
}

int main(void) {
	scanf("%d", &n);

	for (int i = 1; i <= n; i++)
		scanf("%lld", input + i);

	if (n % 2 == 0)
		printf("%lld\n", func(1, 1));
	else
		printf("%lld\n", func(2, 1));
}