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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
#define chmax(x,y) x = max(x,y)
#define chmin(x,y) x = min(x,y)

ll solve(int n,int k,vector<int>a){
    ll dp1[n][k+1];
    ll dp2[n][k+1];
    memset(dp1,0,sizeof(dp1));
    memset(dp2,0,sizeof(dp2));
    int mx=-1e9;
    rep(i,n){
        dp1[i][1]=a[i];
        dp2[i][1]=mx;
        mx=max(mx,a[i]);
        }
    for(int i=1;i<n;i++)
     for(int j=0;j<=k;j++){
       if(j>(i+1)/2)continue;
       if(j==0)continue;
       dp1[i][j]=dp2[i-1][j-1]+a[i];
       dp2[i][j]=max(dp1[i-1][j],dp2[i-1][j]);
    }
    return max(dp1[n-1][k],dp2[n-1][k]);
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);

int n;
cin>>n;
int k=n/2;

vector<int>a(n);
rep(i,n)cin>>a[i];

cout<<solve(n,k,a)<<endl;
return 0;
}