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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>

using namespace std ;

int main()
{
    int N ;
    while(cin >> N)
    {
        int ary[N],t1 = 0,t2 = 0,t3 = 0,t4 = 0 ;
        for(int i = 0;i < N;i++)
        {
            cin >> ary[i] ;
        }

        for(int i = 1;i < (N-1);i++)
        {
            if(( ( (ary[i-1])*(-1) ) + ( (ary[i])*(-1) ) ) > ( ( (ary[i])*(-1) ) + ( (ary[i+1])*(-1) ) ) )
            {
                t1 = ((ary[i-1])*(-1) ) + ( (ary[i])*(-1) ) ;
            }
            else
            {
                t2 = ((ary[i])*(-1) ) + ( (ary[i+1])*(-1) ) ;
            }

            if(( (ary[i-1])  + (ary[i]) ) > ( (ary[i]) +  (ary[i+1]) ) )
            {
                t3 = (ary[i-1])  + (ary[i]) ;
            }
            else
            {
                t4 = (ary[i]) +  (ary[i+1]) ;
            }


            if(t1 > t2 && t1 > t3 && t1 > t4)
            {
                ary[i-1] = (ary[i-1])*(-1) ;
                ary[i] = (ary[i])*(-1) ;
            }
            else if(t2 > t1 && t2 > t3 && t2 > t4)
            {
                ary[i] = (ary[i])*(-1) ;
                ary[i+1] = (ary[i+1])*(-1) ;
            }
            else if(t3 > t1 && t3 > t2 && t3 > t4)
            {
                ary[i-1] = ary[i-1] ;
                ary[i] = (ary[i]) ;
            }
            else if(t4 > t1 && t4 > t2 && t4 > t3)
            {
                ary[i] = ary[i] ;
                ary[i+1] = (ary[i+1]) ;
            }
        }
        
        int ans = 0 ;
        for(int i = 0;i < N;i++)
        {
            ans = ans+ary[i] ;
        }
        cout << ans ;
    }
}