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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
public class Main{
    public static void main(String[] args) throws IOException{
        Reader r = new Reader();        
        int N = r.i();
        long[] A = r.ll();
        long sum = 0;
        for(int i = 0; i < N-1; i++){
            if(A[i] < 0){
                A[i] = -A[i];
                A[i+1] = -A[i+1];
            }
            sum += A[i];
        }
        if(A[N-1] < 0){
            A[N-1] = -A[N-1];
            sum -= 2*min(A);
        }
        sum += A[N-1];
        print(sum);
    }
    static void print(Object o){
        System.out.println(o.toString());
    }
    static int max(int[] ii){
        int max = ii[0];
        for(int i : ii) if(max < i) max = i;
        return max;
    }
    static int min(int[] ii){
        int min = ii[0];
        for(int i : ii) if(i < min) min = i;
        return min;
    }
    static long max(long[] ll){
        long max = ll[0];
        for(long l : ll) if(max < l) max = l;     
        return max;
    }
    static long min(long[] ll){
        long min = ll[0];
        for(long l : ll) if(l < min) min = l;
        return min;
    }
    static double max(double[] dd){
        double max = dd[0];
        for(double d : dd) if(max < d) max = d;
        return max;
    }
    static double min(double[] dd){
        double min = dd[0];
        for(double d : dd) if(d < min) min = d;
        return min;
    }
    static class Reader extends BufferedReader{
        Reader(){
            super(new InputStreamReader(System.in));
        }
        String s() throws IOException{
            return readLine();
        }
        String[] ss() throws IOException{
            return readLine().split(" ");
        }
        int i() throws IOException{
            return Integer.parseInt(readLine());
        }
        int[] ii() throws IOException{
            String[] ss = ss();
            int size = ss.length;
            int[] ii = new int[size];
            for(int j = 0; j < size; j++) ii[j] = Integer.parseInt(ss[j]);
            return ii;
        }
        int[] ii(int N) throws IOException{
            int[] ii = new int[N];
            for(int j = 0; j < N; j++) ii[j] = i();
            return ii;
        }
        long l() throws IOException{
            return Long.parseLong(readLine());
        }
        long[] ll() throws IOException{
            String[] ss = ss();
            int size = ss.length;
            long[] ll = new long[size];
            for(int j = 0; j < size; j++) ll[j] = Long.parseLong(ss[j]);            
            return ll;
        }
        long[] ll(int N) throws IOException{
            long[] ll = new long[N];
            for(int j = 0; j < N; j++) ll[j] = i();            
            return ll;
        }
    }
}