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 Pascal (FPC 3.0.4) (judge verdict: Accepted)

var
	N,i,a,j,ni,nj:Longint;
	dp:Array[0..200000,0..3]of int64;
begin
read(N);
for i:=0 to N do for j:=0 to 2 do dp[i,j]:=-1000000000000000000;
dp[0,0]:=0;
for i:=0 to N-1 do begin
read(a);
for j:=0 to 2 do begin
if dp[i+1,j+1]<dp[i,j]then dp[i+1,j+1]:=dp[i,j];
ni:=i+2;if ni>N then ni:=N;
nj:=j;if i+2=N then inc(nj);
if dp[ni,nj]<dp[i,j]+a then dp[ni,nj]:=dp[i,j]+a;
end;
end;
writeln(dp[N,1+N mod 2]);
end.