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 = list(map(int, input().split()))
k = 1+N%2 # 余分なxを入れる個数
INF = 10**18
# dp: i番目まで見たときに、j個の余分なxを入れた時の、和の最大値
dp = [[-INF for _ in range(4)] for _ in range(202020)]
dp[0][0] = 0
for i in range(N):
    for j in range(k+1):
        dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j]) # xを一つ挿入する=>今見てる数字をスキップする
        now = dp[i][j]
        if (i+j)%2==0:
            now += A[i]
        dp[i+1][j] = max(dp[i+1][j], now)
print(dp[N][k])