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 Python (3.8.2) (judge verdict: Wrong Answer)

n = int(input())
arr = list(map(int,input().split()))



if n % 2 == 0:
    tp = sum(arr[::2])
    print(max(sum(arr)-tp,tp))
elif n == 3:
    print(arr[0]+arr[2])
else:
    hoge = -(10**12)
    buf = [[hoge for i in range(3)] for j in range(n)]
    buf[0][0] = arr[0]
    buf[1][1] = arr[1]
    buf[2][2] = arr[2]
    buf[2][0] = arr[0]+arr[2]

    for i in range(3,n):
        buf[i][0] = buf[i-2][0] + arr[i]
        buf[i][1] = max(buf[i-3][0] + arr[i],buf[i-2][1]+arr[i])
        if i >= 4:
            buf[i][2] = max([buf[i-3][1] + arr[i],buf[i-4][0] + arr[i],buf[i-2][2]+arr[i]])
        else:
            buf[i][2] = max([buf[i-3][1] + arr[i],buf[i-2][2]+arr[i]])


    print(max(buf[-1][2],buf[-2][1],buf[-3][0]))