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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

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

typedef long long ll;

const int INF = 1 << 30;
const ll LLINF = 1LL << 62;

int mod = 1000000007;

int main(void){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int N; cin >> N;
    ll A[N]; rep(i, N) cin >> A[i];
    bool zero = false;
    int numneg = 0;
    ll minabs = mod;
    ll acc = 0;
    rep(i, N){
        if(A[i] == 0) zero = true;
        if(A[i] < 0) numneg++;
        if(minabs > abs(A[i])) minabs = abs(A[i]);
        acc += abs(A[i]);
    }
    if(zero) cout << acc << endl;
    else{
        if(numneg%2){
            cout << acc - minabs*2 << endl;
        }
        else cout << acc << endl;
    }
    return 0;
}