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 Perl (5.26.1) (judge verdict: Wrong Answer)

use List::Util qw(max);
my $N= <STDIN>;
my $s= <STDIN>;
my @dat = split(/ /,$s);
my @dp1;#左つめ
my @dp2;#スキップ1
my @dp3;#スキップ2
if($N==1){
    print $dat[0];exit;
}

my $M=  int($N/2);
$dp1[0]=$dat[0];
$dp2[1]=$dat[1];
$dp3[2]=$dat[2];
for(my $i=0;$i<$N;$i++){
  $dp1[$i+2]= $dp1[$i]+$dat[$i+2];
  $dp2[$i+3]= max($dp2[$i+1],$dp1[$i]) + $dat[$i+3];
  $dp3[$i+4]= max($dp3[$i+2],$dp2[$i+1],$dp1[$i]) + $dat[$i+4];
}
if($N==3){
    print $dp1[2];exit;
}

if($N%2){
    print $dp3[$N-1];
}else{
    print $dp2[$N-1];
}
#print "$dp1[$N-1],$dp2[$N-1],$dp3[$N-1]\n";