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;

// macro
#define rep(i,n) for(i=0;i<n;i++)
#define ll long long
#define all(v) v.begin(), v.end()

// code starts
#define MIN -1000000000000000

int main()
{
	int n;cin>>n;
	vector<int> a(n);
	int i,j,k;
	rep(i,n) cin>>a[i];
	vector<vector<ll>> dp(n+1,vector<ll>(3,MIN));
	rep(i,3) dp[0][i]=0;
	rep(j,3)
	{
		for(i=j+1;i<=n;i++)
		{
			if(i==j+1) dp[i][j]=a[i-1];
			else if((i-j-1)%2==0)
			{
				rep(k,j+1)
				{
					dp[i][j]=max(dp[i][j],dp[i-(2+j-k)][k]);
				}
				dp[i][j]+=a[i-1];
			}
			else dp[i][j]=dp[i-1][j];
		}
	}
	dp[n][0]=dp[n-1][0];
	ll ans=MIN;
	rep(i,2+n%2)ans=max(ans,dp[n][i]);
	cout<<ans<<endl;
}
