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 Rust (1.42.0) (judge verdict: Accepted)

use itertools::Itertools;
use proconio::input;
use proconio::marker::{Chars, Usize1};
use std::cmp::*;
use std::collections::*;
use std::iter::Iterator;

#[allow(unused_macros)]
macro_rules! max {
    ($x:expr) => {
      $x
    };
    ($x:expr, $($xs:tt)+) => {
      max($x,max!($($xs)+))
    };
  }

#[allow(unused_macros)]
macro_rules! min {
    ($x:expr) => {
      $x
    };
    ($x:expr, $($xs:tt)+) => {
      min($x,min!($($xs)+))
    };
  }

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

fn main() {
    input! {
    n: usize,
    mut a: [i64; n],
    }

    let mut ans: u64 = 0;

    let a_un_cnt = a.iter().filter(|x| **x < 0).count();
    let min = a.iter().map(|x| x.abs()).min().unwrap();
    let mut minflg = false;
    let mut minnum =0;

    if a_un_cnt % 2 == 0 {
        for i in 0..n {
            if a[i] < 0 {
                ans += a[i].abs() as u64;
            } else {
                ans += a[i] as u64;
            }
        }
    } else {
        for i in 0..n {
            if a[i].abs() == min && !minflg {
                minnum = a[i].abs() as u64;
                minflg = true;
            } else {
                if a[i] < 0 {
                    ans += a[i].abs() as u64;
                } else {
                    ans += a[i] as u64;
                }
            }
        }

        ans = ans - minnum;
    }

    println!("{}", ans);
}