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: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cmath>
#include <vector>
#define rep(i, n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;

int main() {
    const ll INF = 1e16;

    int n;
    cin >> n;
    ll a[n+1];
    for(int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    
    ll dp[n+1][2];  /* dp[i][j] : 先頭からa[i]までを確定し、
                        j = 0 ならa[i+1]はそのまま(iを選ばず操作しない)
                        j = 1 ならa[i+1]の符号が反転している状態(iを選んで操作する)
                        のときのa[i]までの総和の最大値 */

    rep(i, n+1) {
        rep(j, 2) {
            dp[i][j] = 0;
        }
    }
    dp[0][1] = -INF;

    rep(i, n) {
        rep(j, 2) {
            if(j == 0) {
                dp[i+1][j] = max(dp[i][0] + a[i], dp[i][1] - a[i]);
            }
            else if(j == 1) {
                dp[i+1][j] = max(dp[i][0] - a[i], dp[i][1] + a[i]);
            }
        }
    }
    
    cout << dp[n][0] << endl;
    return 0;
}