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;
#define ll long long 
const int N = 2e5 + 10;
ll c[N], sum[N], dp[N];
int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%lld", &c[i]);
	sum[1] = c[1];
	for (int i = 3; i <= n; i += 2)
		sum[i] = sum[i - 2] + c[i];
	dp[1] = 0;
	for (int i = 2; i <= n; i++) {
		if (i & 1)
			dp[i] = max(dp[i - 1], dp[i - 2] + c[i]);
		else 
			dp[i] = max(dp[i - 2] + c[i], sum[i - 1]);
	}
	printf("%lld", dp[n]);
}
/*
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
*/