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: Runtime Error)

package sample;

import java.util.Scanner;

public class Main
{

	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);

		int n = Integer.parseInt(sc.next());

		long[] value = new long[n];

		for (int i = 0; i < n; i++) {
			value[i] = Long.parseLong(sc.next());
		}

		boolean minuscount = false;
		long max = 0;

		for (int i = 0; i < n; i++) {
			if (value[i] < 0) minuscount = !minuscount;
		}

		for (int i = 0; i < n; i++) {
			max += Math.abs(value[i]);
		}

		if (minuscount) {
			long min = Math.abs(value[0]);

			for (int i = 0; i < n; i++) {
				if (min > Math.abs(value[i])) {
					min = Math.abs(value[i]);
				}
			}

			max -= min * 2L;
		}

		System.out.println(max);


		sc.close();
	}

}