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;
#define rep(i,n) for(int i = 0;i<n;i++)
#define erep(i,n) for(int i = 0;i<=n;i++)
#define rep1(i,n) for(int i = 1;i<n;i++)
#define erep1(i,n) for(int i = 1;i<=n;i++)
typedef long long ll;
#define vint vector<int>
#define vvint vector<vector<int>>
#define vstring vector<string>
#define vdouble vector<double>
#define vll vector<ll>:
#define vbool vector<bool>
#define INF 1101010101010101010
#define MOD 1000000007
#define int long long
using P = pair<int,int>;

signed main(){
  int n;
  cin >> n;
  vint A(n);
  int minus = 0;
  int zero = 0;
  int MIN = INF;
  int sum;
  rep(i,n){
    cin >> A[i];
    if(A[i] < 0) minus++;
    if(A[i] == 0) zero++;
    MIN = min(MIN,abs(A[i]));
    sum += abs(A[i]);
  }
  if(minus % 2 == 1 && zero == 0){
    cout << sum - MIN*2 << endl;
  }
  else{
    cout << sum << endl;
  }
}