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;
const ll INF = 1LL << 60;
#define tl 1'000'000'007
#define REP(i, n) for (ll i = 0; i < n; i++)
#define REREP(i, n) for (ll i = n; i >= 0; i--)
#define FOR(i, a, n) for (ll i = a; i < n; i++)
#define REFOR(i, n, a) for (ll i = n; i >= a; i--)
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false); 
    ll n;
    cin >> n;
    vector<ll> A(n);
    int f= 0;
    ll min= INF;
    REP(i,n){
        cin >> A[i];
        
        if(A[i]<0) f^= 1;
        A[i]= abs(A[i]);
        if(min>A[i]) min= A[i];
    }
    ll sum= accumulate(A.begin(), A.end(), 0LL);
    // cout << f << endl;
    cout << ((f)?sum-2*min:sum) << endl;
}