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>
using namespace std;
using ll=long long;
const ll MOD=1000000007;
const double PI=3.14159265358979;
const ll INF= pow(10,18);
typedef pair<ll,ll> P;
typedef vector<ll> vl;
typedef vector<vl> vvl;
#define for(i,a,b) for(ll i=a;i<b;i++)
string abc="abcdefghijklmnopqrstuvwxyz";
string ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ";

int main() {
  ll n;
  cin >> n;
  vl a(n);
  for(i,0,n){
    cin >> a[i];
  }
  vvl dp(n,vl(3));
  for(i,0,n){
    for(j,0,2){
      dp[i][j]=0;
    }
  }
  dp[0][0]=a[0];
  dp[1][1]=a[1];
  dp[2][2]=a[2];
  dp[3][1]=max(dp[0][0],dp[1][1])+a[3];
  dp[2][0]=a[0]+a[2];
  for(i,4,n){
    if(i%2==0){
    dp[i][0]=dp[i-2][0]+a[i];
    }
    dp[i][1]=max(dp[i-2][1]+a[i],a[i]+dp[i-3][0]);
    dp[i][2]=max(dp[i-2][2]+a[i],max(dp[i-3][1],dp[i-4][0])+a[i]);
  }
  if(n%2==0){
    cout << max(dp[n-2][0],dp[n-1][1]) << endl;
  }
  else
    cout << max(max(dp[n-1][2],dp[n-2][1]),dp[n-3][0]) << endl;
}