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>
using namespace std;
typedef long long ll;
#define nl "\n"
#define rep(i,n) for(ll i=0; i<(n); i++)
typedef vector<ll> vl;
typedef vector<vector<ll>> matrix;//matrix A(n, vl(m)); //n*m行列
typedef vector<string> vs;
typedef pair<ll,ll> pl;
typedef vector<pl> vp;
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define sortup(V) sort(V.begin(), V.end())
#define sortdown(V) sort(V.rbegin(), V.rend())
template<class T>bool chmax(T &a, const T &b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T &a, const T &b){if(b<a){a=b;return 1;}return 0;}
#define MOD 1000000007 //998244353


int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    // solution comes here

    ll N;
    cin >> N;
    ll A, ans=0, MIN=1e15, neg=0;
    bool haveZERO=false;
    rep(i, N) {
        cin >> A;
        if(0==A) haveZERO = true;
        neg += (A<0);
        A = abs(A);
        ans += A;
        chmin(MIN, A);
    }
    
    if(haveZERO or neg%2==0) cout << ans << nl;
    else                   cout << ans - 2*MIN << nl;
}