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 PHP (5.6.30) (judge verdict: Accepted)

<?php
//前提条件$size>=2
$size = fgets(STDIN);
$v = fgets(STDIN);
//$vを配列にする
$array = explode(" ", $v);

//配列内に0があれば出力は配列絶対値の和
if(in_array(0,$array)){
  $allPlus = true;
}else{
  $allPlus = false;
}

$arrayAbs = [];
if($allPlus){
  foreach($array as $n){
    if($n < 0){
      $n = abs($n);
    }
    $arrayAbs[] = $n;
  }
  echo array_sum($arrayAbs);
//配列内のマイナスの数が偶数なら出力は絶対値合計、奇数なら絶対値最小値1つのみマイナス
}else{
  $min = abs($array[0]);
  $minusCounter = 0;
  foreach($array as $n){
    if($n < 0){
      $n = abs($n);
      $minusCounter++;
    }
    //if文の判定が先に行われ、その後上記の$n=abs($n)が実行されるため下記でもう一度$absが必要
    if($min > abs($n)){
      $min = $n;
    }
    $arrayAbs[] = $n;
  }
  if($minusCounter%2 === 0){
    echo array_sum($arrayAbs);
  }else{
    //最小値のみ引くことと、全ての和から最小値を2回引くことは同義
    echo array_sum($arrayAbs)-2*$min;
  }
}