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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
 
#define froop1(a,n) for(int i = a;i < n;i++)
#define froop2(a,n) for(int i = a;i > n;i--)
 
#define ll long long
 
#define pq(name,type) priority_queue<type> name
#define vec(name,type,length,value) vector<type> name(length,value)
#define vec2(name,type,length,width,value) vector<vector<type>> name(length,vector<type>(width,value));
 
int main(){
	ll N,count1 = 0,sum = 0,min = 99999999999;
  cin >> N;
  vec(a,ll,N,0);
  froop1(0,N){
   cin >> a.at(i); 
    if(a.at(i) < 0){
      count1++;
    }
    sum += abs(a.at(i));
    if(abs(a.at(i)) < min){
      min = abs(a.at(i));
    }
  }
  if(count1 % 2 == 0){
    cout << sum << endl;
  }else{
    cout << sum - 2*min << endl;
  }
}