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 Fortran (GNU Fortran 9.2.1) (judge verdict: Wrong Answer)

program select_half
  implicit none
  integer(8), parameter :: inf = 1e17
  integer :: n, i, j
  integer(8) :: x = -inf
  integer(8), dimension(-1:200002) :: a = 0_8, l = 0_8, r = 0_8
  integer(8), dimension(-1:200002) :: lm = inf, rm = inf
  logical :: is_even = .false.
  read(*,*) n
  read(*,*) a(1:n)
  do i = 1, n
    l(i) = l(i-2)+a(i)
    lm(i) = min(lm(i-2),a(i))
  end do
  do i = n, 1, -1
    r(i) = r(i+2)+a(i)
    rm(i) = min(rm(i+2),a(i))
  end do
  lm(-1:0) = 0_8
  lm(n+1:n+2) = 0_8
  rm(-1:0) = 0_8
  rm(n+1:n+2) = 0_8
  is_even = mod(n,2) == 0
  do i = 0, n
    x = max(x,l(i-1)+r(i+2))
    x = max(x,l(i)+r(i+2))
    x = max(x,l(i-1)+r(i+1))
    if (is_even) cycle
    x = max(x,l(i)+r(i+1)-a(i))
    x = max(x,l(i)+r(i+1)-a(i+1))
    if (mod(i,2) == 1) cycle
    do j = i+4, n, 2
      x = max(x,l(i-1)+r(i+2)-r(j)+r(j+1))
    end do
  end do
  write(*,'(i0)') x
end program select_half