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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

#define ABS(a) (((a) < 0) ? -(a) : (a) )

void solve(long *arr, int len) {
  long sum = 0, min = 1e9 + 100; 
  int i, n_cnt = 0;
  for(i = 0; i < len; ++i) {
    sum += ABS(arr[i]);
    if(arr[i] < 0) { ++n_cnt; }
    if(ABS(arr[i]) < min) { min = ABS(arr[i]); }
  }
  if(n_cnt % 2 != 0) { sum -= (2 * min); }
  printf("%ld\n", sum);
}

int main()
{
  int n, i;
  scanf("%d", &n);
  long *a = (long*)malloc((n+4)*sizeof(long));
  for(i = 0; i < n; ++i) { scanf("%ld", &a[i]); }
  solve(a, n);
  return 0;
}