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 <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

const long long INF = 1LL << 60;

typedef long long ll;

ll max(ll a, ll b){
    if(a<b) return b;
    else return a;
}

int main(){
    ll N=0;
    scanf("%lld",&N);
    ll A[N],DP[N+1][N];
    DP[0][0]=0;DP[1][0]=0;DP[2][0]=0;DP[3][0]=0;
    for(int j = 2;j<=N/2;j++) DP[2*j-2][j] = -INF;
    for(int i = 1;i<N+1;i++) {
        scanf("%lld",A+i-1);
        DP[1][1] = A[0];
        for(int j = i/2-1;j<=(i+1)/2&&j<=N/2;j++) {
            if(i == 1)break;
            if(j<=0) continue;
            DP[i][j] = max(DP[i-2][j-1]+A[i-1],DP[i-1][j]);
        }
    }
    printf("%lld\n",DP[N][N/2]);
    return 0;
}