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>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define DEBUG
#define PI 3.141592653589793238462643383279

#define _GLIBCXX_DEBUG
#ifdef DEBUG
#define s(...) show(__VA_ARGS__);
#define sl(...) show(__VA_ARGS__);cout<<endl;
#else
#define s(...)
#define sl(...)
#endif
void show() {}
template <class Head,class... Args>void show(Head t,Args... args){std::cout<<t<<" ";show(args...);}
#define REP(i, n) for(ll i = 1; i <= n; i++)
#define REPR(i, n) for(ll i = n; i >= 1; i--)
#define FOR(i, m, n) for(ll i = m; i <= n; i++)
const ld INF = 1e9+123;
#define VEC(type,A,N) vector<type> A(N+10)
#define VECi(type,A,N,i) vector<type> A(N+10,i)
#define VEC2(type,A,N,M) vector<vector<type>> A(N+10,vector<type>(M+10))
#define VEC2i(type,A,N,M,i) vector<vector<type>> A(N+10,vector<type>(M+10,i))
#define ALL(v) v.begin(), v.end()
ll frac(ll k){
    int s = 1;
    for (ll i=1; i<=k; ++i)s *= i;
    return s;
}
ll gcd(ll a, ll b){
   if (a%b==0)return(b);
   else return(gcd(b, a%b));
}
ll lcm(ll a,ll b){
    return a*b/gcd(a,b);
}
ll antidiv(ll N,ll C,ll D){
    return N-floor(N/C)-floor(N/D)+floor(N/lcm(C,D));
}
vector<ll> yakusuu(ll N){
    vector<ll> ret;
    for (ll i = 1; i * i <= N; i++) {
        if (N % i == 0) {
            ret.push_back(i);
            if (i * i != N) ret.push_back(N / i);
        }
    }
    sort(ret.begin(), ret.end());
    return ret;
}
int main(){
    int N;
    cin>>N;
    vector<ll> A(N);
    int cnt=0;
    ll sum=0;
    FOR(i,0,N-1){
        cin>>A[i];
        if(A[i]<0){
            cnt++;
            A[i]*=-1;
        }
        sum+=A[i];
    }
    
    sort(ALL(A));
    if(cnt%2==1)sum-=A[0]*2;
    cout<<sum;
    //cout<<fixed<<setprecision(32)<<m;
    //sort(ALL(C),[](auto& a, auto& b) {return a[2] < b[2];});
    //if(equal(D.begin(), D.end()-1, K.begin()))cout<<"Yes";else cout<<"No";
    cout<<endl;
    return 0;
}