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>
#define LL long long
using namespace std;
const int maxn=200005;
int n,m=1,val[maxn];
LL f[maxn][2][3];
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&val[i]);
	if (n&1) m++;
	f[1][1][0]=val[1];
	for (int i=2;i<=n;i++)
		for (int j=0;j<=m;j++)
		{
			f[i][0][j]=f[i-1][1][j];
			f[i][1][j]=f[i-1][0][j]+val[i];
			if (j) f[i][0][j]=max(f[i][0][j],f[i-1][0][j-1]);
		}
	printf("%lld\n",max(f[n][0][m-1],f[n][1][m]));
	return 0;
}
