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(void){
    // Your code here!
    int N;
    long long int A[200010];
    int i=0;
    scanf("%d",&N);
    for(i=0;i<N;i++)
    {
        scanf("%lld",&A[i]);
    }
    
    long long int ans=0;
    
    i=0;
    int flag=0; //マイナスが偶数(最後プラスになる)とき
    while(i<N)
    {
        if(A[i]<0)
        {
            ans=ans+(A[i]*(-1));
            if(flag==0) flag=1;
            else        flag=0;
        }
        
        if(A[i]>0)
        {
            ans+=A[i];
        }
        
        if(A[i]==0)
        {
            i++;
            if(A[i]<0)
            {
                ans=ans+(A[i]*(-1));
            }
            else ans+=A[i];
            flag=0;
        }
        

        i++;
        
    }
    
    if(flag==0)
    {
        printf("%lld",ans);
    }
    if(flag==1)
    {
        printf("%lld",ans-(A[N-1]*2));
    }
    return 0;
}