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 <math.h>
#include <string.h>
#include <stdlib.h>
int main()
{
    int n;
    scanf("%d",&n);
    int b[100005],i;
    for(i=0;i<n;i++)
        scanf("%d",&b[i]);
    long long sum=0;
    for(i=1;i<n;i++)
    {
        if(b[i]<0&&b[i-1]<0)
        {b[i]=-b[i];b[i-1]=-b[i-1];}
    }
    if(b[0]<=0&&b[1]<=abs(b[0])) {b[0]=-b[0];b[1]=-b[1];}
    for(i=1;i<n-1;i++)
    {
        if(b[i]<0)
        {
            long long sum1=b[i]+b[i+1];
      if(sum1==-sum1) {b[i]=-b[i];b[i+1]=-b[i+1];}
            else if(b[i-1]<=abs(b[i])&&b[i-1]<=b[i+1])
            {b[i]=-b[i];b[i-1]=-b[i-1];}
            else if(b[i+1]<=abs(b[i])&&b[i+1]<=b[i-1])
                {b[i]=-b[i];b[i+1]=-b[i+1];}
        }

    }
    for(i=0;i<n;i++)
        sum+=b[i];
    printf("%lld\n",sum);
    return 0;
}