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<math.h>
#include<stdlib.h>

int main(){
  int i;
  int count = 0;
  int n;
  int low_i;
  int low;
  long long int ans = 0;
  long long int a[100000];
  long long int b[100000];

  scanf("%d", &n);
  for(i = 0; i < n; i++){
    scanf("%lld", &a[i]);
  }

  low = llabs(a[0]);
  low_i = 0;

  for(i = 0; i < n; i++){
    if(a[i] < 0){
      b[i] = - a[i];
      count++;
    }
    else{
      b[i] = a[i];
    }

    if(low > b[i]){
      low = b[i];
      low_i = i;
    }
  }

  for(i = 0; i < n; i++){
    if(i == low_i &&  (count = count % 2) != 0){
      ans +=  - b[i];
    }
    else {
      ans += b[i];
    }
  }

  printf("%lld\n", ans);


  return 0;
}