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)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
template<typename T> bool chmax(T &a,T b){if(a<b){a=b;return true;}return false;}
template<typename T> bool chmin(T &a,T b){if(a>b){a=b;return true;}return false;}
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(vec) vec.begin(),vec.end()
typedef long long ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
typedef tuple<ll,ll,ll> tlll;
typedef tuple<int,int,int> tiii;
const ll mod=1e9+7;
const int inf=1<<30;

int main(){
  int n; cin >> n;
  vector<ll> a(n);
  int hu=0;
  ll sum=0;
  rep(i,n){
    cin >> a[i];
    if(a[i]<0){
      hu++;
      a[i]=-a[i];
    }
    sum+=a[i];
  }
  sort(all(a));
  if(hu%2==0){
    cout << sum << endl;
  }
  else{
    cout << sum-a[0] << endl;
  }
  
}