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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val aa = readLine()!!.split(" ").map { it.toLong() }.toMutableList()

    for (i in 0 until n - 1) {
        val f = aa[i]
        val s = aa[i + 1]

        if (f > 0 && s > 0) {

        } else if (f > 0 && s < 0) {

        } else {
            aa[i] = f * -1
            aa[i + 1] = s * -1
        }

    }

    if (aa.all { it > 0 }) {
        println(aa.sum())
    } else {
        var min = Long.MAX_VALUE
        var sum = 0L
        for (i in 0 until n) {
            min = Math.min(min, Math.abs(aa[i]))
            sum += Math.abs(aa[i])
        }

        println(sum - min * 2)
    }
}