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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#include <ctype.h>

int main(int argc, void **argv)
{
	int N;
	long long A[100000];
	unsigned long long sum = 0;
	int count = 0;

	scanf("%d", &N);
	for (int i = 0; i < N; i++)
	{
		scanf("%lld", &A[i]);
		if( A[i] < 0 ) count++;
	}

	if( count % 2 == 0 )
	{
		for (int i = 0; i < N; i++)
		{
			unsigned long long abs;
			abs = A[i] >= 0 ? A[i] : -A[i];
			sum += abs;
		}
	}
	else
	{
		unsigned long long min = 1 * 1000 * 1000 * 1000;

		for (int i = 0; i < N; i++)
		{
			unsigned long long abs;
			abs = A[i] >= 0 ? A[i] : A[i] * -1;
			sum += abs;
			min = min > abs ? abs : min;
		}

		sum -= min * 2;
	}

	printf("%lld\n", sum);

	return 0;
}