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.IOException;
import java.io.InputStream;
import java.util.NoSuchElementException;

class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte(){
        if(ptr < buflen){
            return true;
        }else{
            ptr = 0;
            try{
                buflen = in.read(buffer);
            }catch(IOException e){
                e.printStackTrace();
            }
            if(buflen <=0){
                return false;
            }
        }
        return true;
    }

    private int readByte(){
        if(hasNextByte())return buffer[ptr++];
        else return -1;
    }

    private static boolean isPrintableChar(int c){
        return 33<=c && c<=126;
    }
    public boolean hasNext(){
        while(hasNextByte() && !isPrintableChar(buffer[ptr]))ptr++;
        return hasNextByte();
    }

    public String next(){
        if(!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)){
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public long nextLong(){
        if(!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if(b == '-'){
            minus = true;
            b = readByte();
        }
        if(b < '0' || '9' < b){
            throw new NumberFormatException();
        }
        while(true){
            if('0' <= b && b<='9'){
                n*=10;
                n+=b-'0';
            }else if(b==-1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt(){
        long nl = nextLong();
        if(nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDoutble(){return Double.parseDouble(next());}
}
public class Main {
    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        int n = fs.nextInt();
        int[] a = new int[n];
        for(int i=0;i<n;i++)a[i] = fs.nextInt();

        long sum =0;
        int min = Integer.MAX_VALUE;
        int cnt = 0;
        for(int i=0;i<n;i++){
            if(a[i] < 0)cnt++;
            sum+=Math.abs(a[i]);
            min = Math.min(min,Math.abs(a[i]));
        }

        if(cnt%2==0){
            System.out.println(sum);
        }else{
            System.out.println(sum-2*min);
        }
    }
}