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: Wrong Answer)

#include <stdio.h>

#define N 100000

int main() {
	static int aa[N];
	int n, i;
	long long ans;

	scanf("%d", &n);
	for (i = 0; i < n; i++)
		scanf("%d", &aa[i]);
	for (i = 0; i < n - 2; i++)
		if (aa[i] < 0)
			aa[i] = -aa[i], aa[i + 1] = -aa[i + 1];
	if (-aa[n - 2] + -aa[n - 1] > aa[n - 2] + aa[n - 1])
		aa[n - 2] = -aa[n - 2], aa[n - 1] = -aa[n - 1];
	ans = 0;
	for (i = 0; i < n; i++)
		ans += aa[i];
	printf("%lld\n", ans);
	return 0;
}