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())
l=list(map(int,input().split()))
INF=float('inf')
if n<=3:
    print(max(l))
elif n%2==1:
    dp=[[-INF]*3 for i in range(n+1)]
    dp[1][0]=l[0]
    dp[2][1]=l[1]
    dp[3][0]=l[0]+l[2]
    dp[3][2]=l[2]
    for i,j in enumerate(l[3:],4):
        dp[i][0]=dp[i-2][0]+j
        dp[i][1]=max(dp[i-2][1]+j,dp[i-3][0]+j)
        dp[i][2]=max(dp[i-2][2]+j,dp[i-3][1]+j,dp[i-4][0]+j)
    ans=max(dp[-3][0],dp[-2][1],dp[-1][2])
    print(ans)    
else:
    print(max(sum(l[::2]),sum(l[1::2])))