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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

main :: IO ()                                                                                                                                         
main = do                                                                                                                                             
  _ <- getLine                                                                                                                                        
  as <- map read . words <$> getLine :: IO [Integer]                                                                                                  
  let (c,a,b) = solve as 0 0 1000000000                                                                                                               
  print $ if even c then a else a - 2 * b                                                                                                             
                                                                                                                                                      
solve :: [Integer] -> Int -> Integer -> Integer -> (Int,Integer,Integer)                                                                              
solve [] c a b = (c,a,b)                                                                                                                              
solve (x:xs) c a b                                                                                                                                    
  | x >= 0 && x < b = solve xs c (a + x) x                                                                                                            
  | x >= 0 && x >= b = solve xs c (a + x) b                                                                                                           
  | x < 0 && (-x) < b = solve xs (c + 1) (a - x) (-x)                                                                                                 
  | x < 0 && (-x) >= b = solve xs (c + 1) (a - x) b