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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

/* メモリ食い過ぎ注意 */
#define int ll
#define ALL(v) v.begin(), v.end()
#define _REP(i, n) for (int i = 0; i < n; i++)
#define REP(n) _REP(i, n)

/* define const */
const ll INF = (1ll << 60);
const ll MOD = (ll)1e9 + 7;
const ll MAX = 1e6;
/* end defineing */

signed main()
{
  int n;
  ll a[100001];
  ll aa[100001];
  cin >> n;
  for (int i = 0; i < n; i++)
  {
    cin >> a[i];
    aa[i] = a[i];
  }

  int sum = 0;
  for (int i = 0; i < n - 2; i++)
  {
    if (a[i] < 0)
    {
      a[i] *= -1;
      a[i + 1] *= -1;
    }
    sum += a[i];
  }

  int sum1 = a[n - 2] + a[n - 1];
  int sum2 = (a[n - 2] + a[n - 1]) * -1;
  sum = max(sum + sum1, sum + sum2);

  int sum3 = 0;
  for (int i = n - 1; i > 0; i--)
  {
    if (aa[i] < 0)
    {
      aa[i] *= -1;
      aa[i - 1] *= -1;
    }
    sum3 += aa[i];
  }
  int sum4 = aa[0] + aa[1];
  int sum5 = (aa[0] + aa[1]) * -1;

  cout << max(sum, max(sum + sum4, sum + sum5)) << endl;
}