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)

import java.util.*;
import java.util.stream.*;

public class Main {
  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int N = in.nextInt();
    int[] A = new int[N];
    int cnt = 0;
    boolean zero = false;
    for (int i = 0; i < N; i++) {
      A[i] = in.nextInt();
      if (A[i] < 0) {
        cnt++;
      }
      if (A[i] == 0) {
        zero = true;
      }
    }

    long ans = 0;
    for (int i = 0; i < N; i++) {
      ans += Math.abs(A[i]);
    }
    if (cnt % 2 == 0 || zero) {
      System.out.println(ans);
    } else {
      Arrays.sort(A);
      int pos = 0;
      while (A[pos] < 0) {
        pos++;
      }
      System.out.println(ans - Math.abs(A[pos]) * 2L);
    }
  }
}