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: Wrong Answer)

#include<bits/stdc++.h> 
using namespace std;
typedef long long ll;
template<typename T1,typename T2> bool chmin(T1 &a,T2 b){if(a<=b)return 0; a=b; return 1;}
template<typename T1,typename T2> bool chmax(T1 &a,T2 b){if(a>=b)return 0; a=b; return 1;}

ll n,a[222222];
ll ans = -1e18;
void solve(){
    ll sum = 0;
        for(int i=0;i<n;i++) sum += a[i];
        ll dp[n][3];
        for(int i=0;i<n;i++){
            dp[i][0]=1e18;
            dp[i][1]=1e18;
            dp[i][2]=1e18;
        }
        dp[0][0] = a[0];
        dp[1][0] = a[1];
        dp[1][1] = a[1];
        dp[1][2] = a[0]+a[1];
        for(int i=2;i<n;i++){
            dp[i][0] = dp[i-2][0] + a[i];
            if(i%2==0){
                dp[i][1] = min(dp[i-2][1],dp[i-1][0]) + a[i];
            }
            else{
                dp[i][2] = min(dp[i-2][2],dp[i-1][1]) + a[i];
            }
        }
        ans = sum - min(dp[n-1][1],dp[n-2][2]);
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(20);

    cin>>n;
    for(int i=0;i<n;i++) cin>>a[i];
    if(n%2==0){
        ll dp[n]={};
        for(int i=0;i<n;i++) dp[i] = -1e18;
        dp[0]=a[0], dp[1]=a[1];
        for(int i=2;i<n;i++){
            if(i%2){
                dp[i] = max(dp[i-2],dp[i-3])+a[i];
            }
            else{
                dp[i] = dp[i-2] + a[i];
            }
        }
        chmax(ans,max(dp[n-1],dp[n-2]));
    }
    else{
        solve();
        ll dp[n+1][2];
        for(int i=0;i<=n;i++){
            dp[i][0]=-1e18;
            dp[i][1]=-1e18;
        }
        dp[0][0] = a[0];
        dp[1][0] = a[1];
        for(int i=2;i<n;i++){
            dp[i][0] = dp[i-2][0] + a[i];
            if(i%2 && i>=3) chmax(dp[i][0],dp[i-3][0]+a[i]);
            if(i>=3){
                if(i==3) dp[i][1] = a[i];
                ll ret = dp[i-2][1];
                if(i%2==0) chmax(ret,dp[i-3][0]);
                if(i%2 && i>=4) chmax(ret,dp[i-4][0]);
                dp[i][1] = max(dp[i][1],ret + a[i]);
            }
            else dp[i][1] = a[i];
        }
        chmax(ans,max(dp[n-1][1],dp[n-2][0]));
        chmax(ans,dp[n-2][1]);
        if(n>=3) chmax(ans,dp[n-3][0]);
    }
    cout << ans << endl;
}