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 <cstdio>
#include <algorithm>
#define maxn 200010
#define LL long long
using namespace std;
int a[maxn];
LL dp[maxn],sum[maxn];
int main(){
	int n,i;
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d",&a[i]);
	for(i=1;i<=n;i++)
		if(i%2)sum[i]=sum[i-1]+a[i];
		else sum[i]=sum[i-1];
	for(i=2;i<=n;i++)
		if(i&1)dp[i]=max(dp[i-2]+a[i],dp[i-1]);//i是奇数,i/2==(i-1)/2
		else dp[i]=max(dp[i-2]+a[i],sum[i-1]);//i是偶数
	printf("%lld\n",dp[n]);
	return 0;
}