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())
A = list(map(int, input().split()))
inf = float('inf')
ans = -1 *inf

if N % 2 == 0:
    S_odd = [0]*(N//2+1)
    S_even = [0]*(N//2+1)
    s_odd = 0
    s_even = 0
    for i in range(len(A)):
        if i%2 == 0:
            s_odd += A[i]
            S_odd[i//2+1] = s_odd
        else:
            s_even += A[i]
            S_even[i//2+1] = s_even
    
    s = 0
    for j in range(N//2+1):
        s = S_odd[j] + S_even[N//2]-S_even[j] 
        ans = max(ans,s)
else:
    S_odd = [0]*(N//2+2)
    S_even = [0]*(N//2+1)
    odd = []
    even = []
    s_odd = 0
    s_even = 0
    for i in range(len(A)):
        if i%2 == 0:
            s_odd += A[i]
            S_odd[i//2+1] = s_odd
            odd.append(A[i])
        else:
            s_even += A[i]
            S_even[i//2+1] = s_even
            even.append(A[i])
    
    s = 0
    ans = sum(odd)

    for j in range(N//2):
        for k in range(j,N//2):
            s = S_odd[j] + S_even[k]-S_even[j] +S_odd[N//2+1]-S_odd[k+1]
            ans = max(ans,s)

print(ans)