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>

int main(){
  long int N;
  long int A[111111],B[111111];
  scanf("%ld",&N);
  long int i;
  for(i=0;i<=N-1;i++)
    scanf("%ld",&A[i]);
  long int count = 0;
  for(i=0;i<=N-1;i++){
    if(A[i] < 0){
      B[i] = - A[i];
      count++;
      };
    if(A[i] >= 0) B[i] = A[i];
    };
  long int min = B[0];
  for(i=1;i<=N-1;i++)
    if(min > B[i]) min = B[i];
  long int sum = 0;
  for(i=0;i<=N-1;i++)
    sum += B[i];
  if(count % 2 == 1)
    sum = sum - 2 * min;
  printf("%ld",sum);
}