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 <bits/stdc++.h>

using namespace std;

int main(){
	int n;
	scanf("%d", &n);
	int arr[n];
	for(int i = 0; i < n; i++){
		scanf("%d", &arr[i]);
	}	
	long long soma = 0;
	int mini = 1e9;
	for(int i = 0; i < n - 1; i++){
		if(arr[i] < 0){
			arr[i] *= -1;
			arr[i + 1] *= -1;
		}
		mini = min(mini, arr[i]);
		soma += arr[i];
	}
	if(arr[n - 1] < 0){
		if(arr[n - 1] * -1 >= mini){
			soma -= mini;
			soma += -1LL * arr[n - 1];
		}else{
			soma += arr[n - 1];
		}
	}else{
		soma += arr[n - 1];
	}
	printf("%lld\n", soma);
	return 0;
}