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 <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;

int main(int argc, char *argv[]){

  int N;
  cin>>N;
  vector<int> A(N);
  int a;
  int mcount=0;
  int zerocount=0;
  for(int i=0; i<N; i++){
    cin>>a;
    A[i]=a;
    if(a<0) mcount++;
    if(a==0) zerocount++;
  }
  if(mcount%2==0 || zerocount!=0){
    for(int i=0; i<N-1; i++){
      if(A[i]<0 && A[i+1]>=0){
	A[i]*=-1;
	A[i+1]*=-1;
      }
      else if(A[i]<=0 && A[i+1]<=0){
	A[i]*=-1;
	A[i+1]*=-1;
      }
      
      else continue;
    }
    long long sum=0;
    for(int i=0; i<N; i++){
      sum+=A[i];
    }
    cout<<sum<<endl;
    return 0;
  }
  else{
    vector<int> tmp(N);
    for(int i=0; i<N; i++){
      tmp[i]=abs(A[i]);
    }
    int min = *std::min_element(tmp.begin(), tmp.end());
    long long sum=0;
    for(int i=0; i<N; i++){
      sum+=abs(A[i]);
    }
    sum=sum-(2*min);
    cout<<sum<<endl;
  }
}