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: Accepted)

#include <iostream>
#include <map>
#include <numeric>
#include <vector>

using namespace std;

int main(void){
    int n;
    cin >> n;
    vector<int> v;
    vector<bool> vb;
    v.reserve(n);
    vb.reserve(n - 1);
    for(int i = 0; i < n; ++ i){
        int tmp;
        cin >> tmp;
        v.push_back(tmp);
    }
    long tmpmax = 0, sum = 0;
    for(int i = 0; i < n; ++ i) sum += ((v[i] < 0)? -v[i] : v[i]);
    for(int i = 0; i < n; ++ i){
        if(i == 0) for(int j = n - 2; 0 <= j; -- j) vb[j] = ((j == n - 2)? (v[j + 1] < 0) : (vb[j + 1] != (v[j + 1] < 0)));
        else vb[i - 1] = ((i == 1)? v[0] < 0 : vb[i - 2] != v[i - 1] < 0);
        if(((i == 0)? vb[0] : ((i == n - 1)? vb[n - 2] : (vb[i - 1] != vb[i]))) == (v[i] < 0)){
            tmpmax = sum;
            break;
        }else{
            tmpmax = max(tmpmax, sum - 2 * abs(v[i]));
        }
    }
    cout << tmpmax << endl;
    
    return 0;
}