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.

Examples

Input

3
-10 5 -4


Output

19


Input

5
10 -4 -8 -11 3


Output

30


Input

11
-1000000000 1000000000 -1000000000 1000000000 -1000000000 0 1000000000 -1000000000 1000000000 -1000000000 1000000000


Output

10000000000
import java.lang.Math;
import java.util.Scanner;
import java.util.Arrays;
import java.util.Collections;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
        Integer[] A = new Integer[N];

        int numberOfNegatives = 0;
        int minAbs = Integer.MAX_VALUE;

        for (int i = 0; i < N; i++) {
            A[i] = sc.nextInt();
            if (A[i] < 0) { numberOfNegatives++; }
            int abs = Math.abs(A[i]);
            if (abs < minAbs) { minAbs = abs; }
        }

        long answer = 0;
        for (int i = 0; i < N; i++) {
            answer += Math.abs(A[i]);
        }

        // If the number of negatives are even, all negatives could be flipped to positives
        // if odd, one must be retained somewhere in the array
        if (numberOfNegatives % 2 == 1) {
            answer -= (2 * minAbs);
        }

        System.out.println(answer);
    }
}
