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>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define int long long
using namespace std;
typedef pair<int, int> P;
const int mod = 1000000007;
const int INF = 1e18;

signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<int> a(n), c;
    int sum = 0;
    rep(i, 0, n){
        cin >> a[i];
        sum += a[i];
        if(a[i] <= 0){
            c.push_back(-a[i]);
            sum += 2 * (-a[i]);
        }
    }
    if(c.size() % 2){
        int MIN = INF;
        for(int i = 0; i < c.size(); i++){
            MIN = min(MIN, c[i]);
        }
        for(int i = 0; i < n; i++){
            if(a[i] < 0){
                if(i != 0 && a[i - 1] > 0) MIN = min(MIN, a[i - 1]);
                if(i != n - 1 && a[i + 1] > 0) MIN = min(MIN, a[i + 1]);
            }
        }
        // cout << MIN << endl;
        sum -= 2 * MIN;
    }
    cout << sum << endl;
}