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 <string.h>
#include <stdlib.h>
#include <math.h>
#include <limits.h>
long long asc(const void *a, const void *b) {
  long long *A = (long long *)a;
  long long *B = (long long*)b;
  if (*A > *B) return 1;
  if (*A < *B) return -1;
  return 0;
}
long long maxim(long long a, long long b) { return a > b ? a : b; }
int main(){
    long long n,i,sign=1,ans=0,a[100010]={},b[100010]={};
    scanf("%lld",&n);
    for(i=0; i<n; i++){
        scanf("%lld",&a[i]);
    }
    for(i=0; i<n; i++){
        sign*=a[i];
    }
    if(sign>=0){
        for(i=0; i<n; i++){
            ans+=maxim(a[i],-a[i]);
        }
        printf("%lld",ans);
        return 0;
    }
    for(i=0; i<n; i++){
        a[i]=maxim(a[i],-a[i]);
    }
    qsort(a,sizeof(n)/sizeof(long long),sizeof(long long),asc);
    for(i=0; i<n; i++){
        ans+=a[i];
    }
    ans-=2*a[0];
    printf("%lld\n",ans);
    return 0;
}