Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

6
1 2 3 4 5 6

Sample Output 1

12

Choosing 2, 4, and 6 makes the sum 12, which is the maximum possible value.

Sample Input 2

5
-1000 -100 -10 0 10

Sample Output 2

0

Choosing -10 and 10 makes the sum 0, which is the maximum possible value.

Sample Input 3

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 3

5000000000

Watch out for overflow.

Sample Input 4

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

Sample Output 4

295

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair

using namespace std;

const LL MOD=1e9+7;

LL n,a[200010],dp[200010][3];

int main()
{
	cin>>n;
	rep(i,n) scanf("%lld",&a[i]);
	rep(i,200005) rep(j,3) dp[i][j]=-1e18;
	if(n%2==0)
	{
		dp[0][0]=a[0];
		dp[1][1]=a[1];
		rep(i,n-1)
		{
			rep(j,2)
			{
				if(dp[i][j]==-1e18) continue;
				if(i+2<n) dp[i+2][j]=max(dp[i+2][j],dp[i][j]+a[i+2]);
				if(i+3<n&&j==0) dp[i+3][j+1]=max(dp[i+3][j+1],dp[i][j]+a[i+3]);
			}
		}
		cout<<max(dp[n-1][1],dp[n-2][0])<<endl;
	}
	else
	{
		dp[0][0]=a[0];
		dp[1][1]=a[1];
		dp[2][2]=a[2];
		rep(i,n)
		{
			rep(j,3)
			{
				if(dp[i][j]==-1e18) continue;
				if(i+2<n) dp[i+2][j]=max(dp[i+2][j],dp[i][j]+a[i+2]);
				if(i+3<n&&j<2) dp[i+3][j+1]=max(dp[i+3][j+1],dp[i][j]+a[i+3]);
				if(i+4<n&&j==0) dp[i+4][j+2]=max(dp[i+4][j+2],dp[i][j]+a[i+4]);
			}
		}
		cout<<max(max(dp[n-1][2],dp[n-2][1]),dp[n-3][0])<<endl;
	}
	return 0;
}