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: Accepted)

n = int(input())
*A, = map(int, input().split())
inf = 10**18
# DP[i][0]=i番目まででi//2個選んだ最大値
# DP[i][1]=i番目まででi//2+1個選んだ最大値
DP = [[-inf]*2 for i in range(n)]
DP[0][0] = 0
DP[0][1] = A[0]
DP[1][0] = 0
DP[1][1] = max(A[0], A[1])
for i in range(2, n):
    if i % 2 == 0:
        DP[i][0] = max(DP[i-2][0]+A[i], DP[i-1][1])
        DP[i][1] = DP[i-2][1]+A[i]
    else:
        DP[i][0] = max(DP[i-2][0]+A[i], DP[i-1][0])
        DP[i][1] = max(DP[i-2][1]+A[i], DP[i-1][1])
if n % 2 == 0:
    print(DP[n-1][1])
else:
    print(DP[n-1][0])