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>
#include<stdlib.h>
#define A 100005
int main(void){
  int N,i,flag=0;
  int AA[A]={0};
  long int count=0;
  scanf("%d",&N);
  for(i=0;i<N;i++){
    scanf("%d",&AA[i]);
  }
  for(i=0;i<N-1;i++){
    if(AA[i]<0 && AA[i+1]>=0 && AA[i+2]<0){
      AA[i]*=-1;
      AA[i+1]*=-1;
      flag=1;
    }else if(flag==1){
      AA[i]*=-1;
      AA[i+1]*=-1;
      flag=0;
    }else if(AA[i]<0 && AA[i+1]<0){
      AA[i]*=-1;
      AA[i+1]*=-1;
    }else if(AA[i]<0 && abs(AA[i+1])<=abs(AA[i])){
      AA[i]*=-1;
      AA[i+1]*=-1;
    }else{
      continue;
    }
  }
  for(i=0;i<N;i++){
    count+=AA[i];
  }
  printf("%ld\n",count);
  return 0;
}