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<algorithm>
#include<vector>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<limits>
#include<iomanip>
#include<cmath>
#include<climits>

using namespace std;



typedef long long ll;
typedef pair<int,int> P;

const int mod=1e9+7;

int gcd(int a,int b){
    if(b==0){
        return a;
    }else{
        return gcd(b,a%b);
    }
}

int main(){
    ll n;
    cin >> n;
    vector<ll> a(n);
    ll cnt=0,AbsMin=INT_MAX,Sum=0;
    bool exist0=false;
    for(ll i=0;i<n;i++){
        cin >> a[i];
        if(a[i]<0){
            cnt++;
        }else if(a[i]==0){
            exist0=true;
        }
        AbsMin=min(AbsMin,abs(a[i]));
        Sum+=abs(a[i]);
    }  
    if(cnt%2==0){
        cout << Sum << endl;
    }else{
        cout << Sum-2*AbsMin << endl;
    }
    return 0;
}