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.util.*;

public class Main {

    Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        new Main().run();
    }

    void run() {

        int n = sc.nextInt();
        int[] arr = new int[n];
        for (int i=0; i<n; i++) arr[i] = sc.nextInt();

        long[][] dp = new long[n+1][2];
        dp[0][1] = Integer.MIN_VALUE;

        for (int i=1; i<=n; i++) {
            dp[i][0] = Math.max(dp[i-1][0]+arr[i-1], dp[i-1][1]-arr[i-1]);
            dp[i][1] = Math.max(dp[i-1][0]-arr[i-1], dp[i-1][1]+arr[i-1]);
        }

//        System.out.println(Math.max(dp[n][0], dp[n][1]));
        System.out.println(Math.max(dp[n][0], 0));
//        System.out.println(Math.max(0, dp[n][1]));


    }

    int gcd(int m, int n) {
        if (n == 0) return m;
        return gcd(n, m%n);
    }


}

