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;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define REP(i, d, n) for(int i=(d); i<(n); ++i)
#define all(v) v.begin(), v.end()
using ll = long long;
using P = pair<int, int>;
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

int main(){
    int n;
    cin >> n;
    vector<ll> a(n);
    rep(i, n){
        cin >> a[i];
    }
    
    vector<vector<ll>> dp(n, vector<ll>(2, -1e18));
    dp[0][0] = a[0];
    dp[0][1] = -a[0];
    REP(i, 1, n){
        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]);
    }
    cout << dp[n-1][0] << endl;

    return 0;
}

/*
int main() {

    int n;
    cin >> n;
    ll ans = 0;
    vector<ll> a(n);
    rep(i, n){
        cin >> a[i];
        ans += abs(a[i]);
    }

    int minus = 0;
    ll m = 1 << 30;
    rep(i, n){
        if(a[i] < 0){
            minus++;
        }
        if(a[i] == 0){
            minus = 0;
            break;
        }
        m = min(m, abs(a[i]));
    }
    if(minus%2 == 0){
        cout << ans << endl;
    }
    else{
        cout << ans - 2*m << endl;
    }
    return 0;
}
*/