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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main{
    public static void main(String args[]){
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        long a[] = new long[n];
        for (int i = 0; i < n; ++i){
            a[i] = in.nextInt();
        }

        for (int i = 0; i < n-1; ++i){
            if (a[i] < 0){
                a[i] = a[i]*(-1);
                a[i+1] = a[i+1]*(-1);
            }
        }
        if (a[n-1] < 0 && Math.abs(a[n-2]) < Math.abs(a[n-1])){
            a[n-2] = a[n-2] * (-1);
            a[n-1] = a[n-1] * (-1);
        }

        long sum = 0;

        for (int i = 0; i < n; ++i){
            sum += a[i];
        }
        System.out.println(sum);
    }
}