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 Ruby (2.3.3) (judge verdict: Runtime Error)

# https://atcoder.jp/contests/abc125/tasks/abc125_d
n = gets.to_i
$nums = gets.split.map(&:to_i)

DP = [Array.new(10**7, nil), Array.new(10**7, nil)]
DP[0][0] = 0
DP[1][0] = -1 * 10**9


def dp(idx, reversed)
  idx = 0 if idx < 0
  unless DP[reversed][idx].nil?
    return DP[reversed][idx]
  end
  num = $nums[idx - 1]

  res = nil
  if reversed == 1
    res = [dp(idx - 1, 0) - num, dp(idx - 1, 1) + num].max
  else
    res = [dp(idx - 1, 0) + num, dp(idx - 1, 1) - num].max
  end
  # puts [idx, res]
  
  DP[reversed][idx] = res;
  return res;
end

current_max = 0
# (n - 1).times do |idx|
#   no_reversed = nums[idx] + nums[idx + 1]
#   reversed = -1 * no_reversed

#   current_max = [current_max + no_reversed, current_max + reversed].max
# end

puts [dp(n, 0), dp(n, 1)].max