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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <inttypes.h>
#include <string.h>
#include <math.h>

int main (int argc, char const *argv[])
{
    int64_t N, A[(int)1e5 + 1] = {0};
    int64_t ans = 0;

    scanf("%"PRId64, &N);
    for (int i = 0; i < N; i++) {
        scanf("%"PRId64, &A[i]);
    }

    for (int i = 0; i < (N - 1); i++) {
        if (A[i] < 0) {
            if ((A[i + 1] > 0) && (A[i + 2] > 0)) {
                if (-A[i] > A[i + 1]) {
                    ans += -A[i];
                    A[i + 1] = -A[i + 1];
                } else {
                    ans += A[i];
                }
            } else {
                ans += -A[i];
                A[i + 1] = -A[i + 1];
            }
        } else {
            ans += A[i];
        }
    }
    ans += A[N - 1];

    printf("%"PRId64"\n", ans);
}