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: Time Limit Exceeded)

import java.util.*;

public class Main {

	public static long[] X;
	public static long[] Y;
	public static boolean[] cached;

	public static long cache(boolean b, int index, int N, int[] A) {
		if (!cached[index]) {
			if (index == 0) {
				Y[index] = -A[index];
				X[index] = A[index];
				cached[index] = true;
			} else {
				long x1 = cache(false, index - 1, N, A);
				long y1 = cache(true, index - 1, N, A);
				Y[index] = Math.max(-A[index] + x1, A[index] + y1);
				X[index] = Math.max(A[index] + x1, -A[index] + y1);
			}
		}
		if (b) {
			return Y[index];
		} else {
			return X[index];
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] A = new int[N];
		for (int i = 0; i < N; i++) {
			A[i] = sc.nextInt();
		}
		X = new long[N];
		Y = new long[N];
		cached = new boolean[N];

		long x = A[N-1] + cache(false, N - 2, N, A);
		long y = -A[N-1] + cache(true, N - 2, N, A);

		System.out.println(Math.max(x, y));
	}
}