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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

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

if n % 2 == 0:
    b = 0
    c = 0
    for i in range(n // 2):
        b += a[i * 2]
        c += a[i * 2 + 1]
    print(max(b, c))
    exit()

if n == 3:
    ans = min(a[0], a[1])
    print(min(ans, a[2]))
    exit()

dp = [[0,0,0] for _ in range(n)]
dp[0][0] = a[0]
dp[1][1] = a[1]
dp[2][0] = a[0] + a[2]
dp[2][2] = a[2]
dp[3][1] = max(a[0], a[1]) + a[3]

for i in range(4, n):
    if i % 2 == 0:
        dp[i][0] = dp[i-2][0] + a[i]
    dp[i][1] = max(dp[i-3][0], dp[i-2][1]) + a[i]
    temp = max(dp[i-4][0], dp[i-3][1])
    dp[i][2] = max(temp, dp[i-2][2]) + a[i]

ans = max(dp[n-1][2], dp[n-2][1])
print(max(ans, dp[n-3][0]))