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: Wrong Answer)

#include <stdio.h>
int main(){

  double n;
  double a[100001];
  double b[100001];
  double tmp = 0;
  int num = 0;
  int sum = 0;

  scanf("%lf", &n);

  for(int cnt = 0; cnt < n; cnt++){
    scanf("%lf", &a[cnt]);
    if(0 > a[cnt]){
      num++;
      if(num == 1){
	tmp = a[cnt];
      }
      if(a[cnt] > tmp){
	tmp = a[cnt];
      }
    }
  }

  if(num / 2 == 0){
    for(int cnt = 0; cnt < n - 1; cnt++){
      if(a[cnt] < 0){	
	a[cnt] = a[cnt] * (-1);
	a[cnt + 1] = a[cnt + 1] * (-1);
      }
    }
    
  }else{
    for(int cnt = 0; cnt < n; cnt++){
      if(a[cnt] < 0 && a[cnt] != tmp){
	a[cnt] = a[cnt] * (-1);
	a[cnt + 1] = a[cnt + 1] * (-1);
      }
    }
  }
  
  for(int cnt = 0; cnt < n; cnt++){
    sum += a[cnt];
  }

  printf("%d", sum);
  printf("\n");

}