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<string>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<queue>
#include<deque>
#include<regex>
#include<stack>
#include<stdio.h>
#include<vector>
#include<set>
#include<map>
#include<iomanip>

#define rep(i, n) for(int i=0;i<n;i++)

typedef int long long ll;
using namespace std;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
const ll MOD = 1e9 + 7;
static const int MAX = 100;
static const int INF = (1 << 23);

int main() {
    ll n;
    cin>>n;
    vector<ll> v(n);
    ll f=0;
    ll sum=0;
    rep(i,n){

        cin>>v[i];
        if(v[i]<0)f++;
        sum+=abs(v[i]);
        v[i]=abs(v[i]);
    }
    sort(v.begin(),v.end());

    if(f%2==0){
        cout<<sum<<endl;

    }else{
        sum-=v[0]*2;
        cout<<sum<<endl;

    }


    return 0;
}