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 <fstream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <limits>
#include <queue>
#include <iomanip>
#include <set>

template<typename T> bool chmax(T &a,T b){
    if(a<b){
        a=b;
        return true;
    }
    return false;
}

template<typename T> bool chmin(T &a,T b){
    if(a>b){
        a=b;
        return true;
    }
    return false;
}

using namespace std;
using ll = long long int;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
const int MOD=1000000007;
//const int MOD=998244353;
const int INTMAX=1001001000;
const ll LLMAX=1010010010010010000;


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    ll N;
    cin>>N;
    vll A(N);
    for(ll i=0;i<N;i++)cin>>A[i];
    
    vvll dp(N,vll(2));//j=0,not changed
    
    for(ll i=0;i<N;i++){
        if(i==0){
            dp[i][0]=A[i];
            dp[i][1]=-A[i];
        }else if(i==N-1){
            dp[i][0]=max(dp[i-1][0]+A[i],dp[i-1][1]-A[i]);
        }else{
            dp[i][0]=max(dp[i-1][0]+A[i],dp[i-1][1]-A[i]);
            dp[i][1]=max(dp[i-1][0]-A[i],dp[i-1][1]+A[i]);
        }
    }
    //for(int i=0;i<N;i++)printf("%lld,%lld\n",dp[i][0],dp[i][1]);
    
    cout<<dp[N-1][0]<<endl;
    
    return 0;
}