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)

#![allow(non_snake_case)]
use proconio::input;
// use std::collections::HashMap;

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

    // 隣接しているのは問題文の通りに処理できる
    // 隣接していなくてもいける??
    // A B C D
    // A B -C -D
    // A - B C -D
    // -A B C -D

    // つまり、適切に2個選んで符号を反転していくということか
    // 負の数を小さい順に処理していく
    // 負の数が0個になったら終わり
    // 負の数が1個だけ余ったら、最小の正の数と入れ替えて大きくなるなら実施して終了

    let mut As: Vec<i64> = Vec::new();
    for a in tmpAs.iter() {
        As.push(*a)
    }

    As.sort();

    let mut ans: i64 = As.iter().sum();
    for i in (0..N).step_by(2) {
        // 2個ずつでやっていると1個余るけど、この1個は使い途が無いのでは?
        if i == N - 1 {
            break;
        }

        let smaller = As[i];
        let bigger = As[i + 1];

        if smaller < 0 && bigger < 0 {
            ans += smaller.abs() * 2;
            ans += bigger.abs() * 2;
        } else if smaller < 0 && bigger >= 0 && smaller + bigger < 0 {
            ans += smaller.abs() * 2;
            ans -= bigger.abs() * 2;
        }
    }



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