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 Go (1.6) (judge verdict: Accepted)

package main

import (
	"fmt"
	"sort"
	"math"
)

func makeabs(A []int) []int {
	for i := 0; i < len(A); i++ {
		A[i] = int(math.Abs(float64(A[i])))
	}
	return A
}

func sum(A []int) int {
	s := 0
	for _, a := range A {
		s += a
	}
	return s
}

func minAbs(A []int) int {
	s := math.Abs(float64(A[0]))

	for i := 1; i < len(A); i++ {
		if t := math.Abs(float64(A[i])); t < s {
			s = t
		}
	}

	return int(s)
}

func main() {
	var N int
	fmt.Scanf("%d", &N)

	A := make([]int, N)
	for i := 0; i < N; i++ {
		fmt.Scanf("%d", &A[i])
	}

	sort.Sort(sort.Reverse(sort.IntSlice(A)))

	n_neg := 0
	for _, a := range A {
		if a < 0 {
			n_neg++
		}
	}

	if n_neg % 2 == 0 {
		fmt.Println(sum(makeabs(A)))
	} else {
		fmt.Println(sum(makeabs(A)) - 2 * minAbs(A))
	}
}