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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author Apolloliu
 */
 
#include <iostream>
#include <fstream>
 
#include <bits/stdc++.h>
 
using namespace std;
 
#define MEM(a,b)    memset(a, b, sizeof(a))
#define UN(a)       sort(all(a)), a.erase(unique(all(a)), a.end())
#define REV(a)      reverse(all(a))
#define MP(a, b)    make_pair(a, b)
#define PAIR(a, b)  pair<int, int>(a, b)
 
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, int> pii;
 
const int MOD = 1e9 + 7;
const int maxn = 123456;
const double EPS = 1e-8;
const ll inf = 1e18;
 
ll a[maxn], b[maxn];
class DFlippingSigns {
public:
    void solve(std::istream& in, std::ostream& out) {
        int N;
        in>>N;
        ll res = -inf, sum = 0;
        for(int i = 0; i < N; ++i)in>>a[i], sum += a[i], b[i] = a[i];
        res = max(res, sum);
        sort(a, a+N);
        for(int i = 1; i < N; i += 2){
            // if(sum - 2*a[i] - a[i-1]*2 >= res){
            //    a[i] = -a[i];
            //    a[i-1] = -a[i-1];
            sum -= (2*a[i] + 2*a[i-1]);
            res = max(res, sum); 
        }
        out<<res<<endl;
    }
};
 
 
int main() {
    // freopen("input.txt", "r", stdin);
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    DFlippingSigns solver;
    std::istream& in(std::cin);
    std::ostream& out(std::cout);
    solver.solve(in, out);
    return 0;
}