Problem:
Score : 400 points

Problem Statement

There are N integers, A_1, A_2, ..., A_N, arranged in a row in this order.

You can perform the following operation on this integer sequence any number of times:

Operation: Choose an integer i satisfying 1 \leq i \leq N-1. Multiply both A_i and A_{i+1} by -1.

Let B_1, B_2, ..., B_N be the integer sequence after your operations.

Find the maximum possible value of B_1 + B_2 + ... + B_N.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

-10^9 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible value of B_1 + B_2 + ... + B_N.

Sample Input 1

3
-10 5 -4

Sample Output 1

19

If we perform the operation as follows:

Choose 1 as i, which changes the sequence to 10, -5, -4.

Choose 2 as i, which changes the sequence to 10, 5, 4.

we have B_1 = 10, B_2 = 5, B_3 = 4. The sum here, B_1 + B_2 + B_3 = 10 + 5 + 4 = 19, is the maximum possible result.

Sample Input 2

5
10 -4 -8 -11 3

Sample Output 2

30

Sample Input 3

11
-1000000000 1000000000 -1000000000 1000000000 -1000000000 0 1000000000 -1000000000 1000000000 -1000000000 1000000000

Sample Output 3

10000000000

The output may not fit into a 32-bit integer type.

Solution in OCaml (4.02.3) (judge verdict: Accepted)

let () = Scanf.scanf "%d\n" @@ fun n ->
  let as_ = Array.init n @@ fun _ -> Scanf.scanf "%d " @@ fun a -> a in
  let dp = Array.make_matrix (n + 1) 2 0 in
  dp.(0).(0) <- 0;
  dp.(0).(1) <- -12345678901234;
  for i = 0 to n - 1 do
    dp.(i + 1).(0) <- max (dp.(i).(0) + as_.(i)) (dp.(i).(1) - as_.(i));
    dp.(i + 1).(1) <- max (dp.(i).(0) - as_.(i)) (dp.(i).(1) + as_.(i))
  done;
  Printf.printf "%d\n" dp.(n).(0)