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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define forn(i, n) for (int i = 0; i < int(n); ++i)
#define for1(i, n) for (int i = 1; i <= int(n); ++i)
#define ms(a, x) memset(a, x, sizeof(a))
#define F first
#define S second
#define all(x) (x).begin(),(x).end()

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
mt19937 gen(random_device{}());
template<typename... Args> void write(Args... args) { ((cout << args << " "), ...); cout<<endl;}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin>>n;
    vector<int> a(n+1);
    for(int i=1;i<=n;++i) cin>>a[i];
    vector<vector<ll>> dp(n+1,vector<ll>(3,-1e18));
    dp[1]={0,0,0};
    for(int i=2;i<=n;++i){
        if(i&1){
            dp[i][0]=max({dp[i-2][0],dp[i-2][1],dp[i-2][2]})+a[i];
            dp[i][1]=dp[i-1][0];
            dp[i][2]=dp[i-1][1];
        }else{
            dp[i][0]=max({dp[i-1][1]+a[i],dp[i-1][2]+a[i]});
            dp[i][1]=dp[i-1][2]+a[i-1];
        }
    }
    cout<<*max_element(all(dp[n]));
    return 0;
}