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 <bits/stdc++.h>

//{ START
using namespace std;
#define int int64_t
#define rep(i, a, n) for (int i = (a); i < (n); ++i)
#define reps(i, a, n) for (int i = (a); i > (n); --i)
#define arep(i, x) for (auto &&i : (x))
#define irep(i, x) for (auto i = (x).begin(); i != (x).end(); ++i)
//降順はgreater<T>()
#define all(x) (x).begin(), (x).end()
#define rv(s) reverse((s).begin(), (s).end())
//gcd lcmはそのままok
#define gcd(a, b) __gcd(a, b)
#define bits(n) (1 << (n))
#define pcnt(x) __builtin_popcountll(x)
//配列内等要素削除
#define unique(x) (x).erase(unique(x).begin(),(x).end()),(x).end())
#define Fixed(n) fixed << setprecision(n)
//総和
#define sowa(n) ((n) * ((n) + 1)) / 2
#define pb(x, a) (x).push_back(a)
#define fb(x, a) (x).push_front(a)
#define eb(x, a) (x).emplace_back(a)
#define F first
#define S second
#define cauto const auto &
using P = pair<int, int>;
using Graph = vector<vector<P>>;
template <class T> //昇順
using min_heap = priority_queue<T, vector<T>, greater<T>>;
template <class T> //降順
using max_heap = priority_queue<T>;
template <class A, class B>
using umap = unordered_map<A, B>;
template <class A>
using uset = unordered_set<A>;
template <class T>
bool chmax(T &a, const T &b) //最大値更新 返り値はbool
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b) //最小値更新 返り値はbool
{
    if (b < a)
    {
        a = b;
        return 1;
    }
    return 0;
}
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
constexpr int INF = 0x3f3f3f3f;
constexpr int LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr int mod1 = 1e9 + 7;
constexpr int mod2 = 998244353;
//} END

signed main()
{
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    int n, cnt = 0, ans = 0;
    cin >> n;
    vector<int> x(n);

    rep(i, 0, n)
    {
        cin >> x[i];
        if (x[i] < 0)
            ++cnt;
    }

    if (cnt % 2 == 0)
    {
        arep(i, x)
        {
            ans += abs(i);
        }
        cout << ans << '\n';
    }
    else
    {
        cnt = INF;
        arep(i, x)
        {
            ans += abs(i);
            chmin(cnt, abs(i));
        }
        cnt = 2 * cnt;
        cout << ans - cnt << '\n';
    }

    return 0;
}