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 <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const long long maxn=400005,mod=1e9+7;
typedef long long ll;

int n;
ll a[maxn];

ll dp[maxn][3];
ll ans=-(1ll<<62);

int main()
{
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
	int lim;
	if(n%2==1) lim=2;
	else lim=1;
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<=lim;j++)
		{
			dp[i][j]=-(1ll<<62);
			if(i==j+1) dp[i][j]=a[i];
			for(int k=0;k<=j;k++)
			{
				if(i-k-2>=1)
					dp[i][j]=max(dp[i][j],dp[i-k-2][j-k]+a[i]);
			}
			//cout<<"dp "<<i<<' '<<j<<' '<<dp[i][j]<<endl;
			if((lim-j)+i==n)
				ans=max(ans,dp[i][j]);
		}
	}
	cout<<ans<<endl;
	return 0;
}