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: Time Limit Exceeded)

#include <stdio.h>

long A[100000];

long max(long a, long b) {
	return (a > b) ? a : b;
}

long rec(int a, int b) {
	long dmax;

	if (a == 0) {
		if (b == 0) {
			dmax = A[a];
		}
		else {
			dmax = -A[a];
		}
	}
	else {
		if (b == 0) {
			dmax = max(rec(a - 1, 0) + A[a], rec(a - 1, 1) - A[a]);
		}
		else {
			dmax = max(rec(a - 1, 0) - A[a], rec(a - 1, 1) + A[a]);
		}
	}

	return dmax;
}

int main() {
	int i, N;
	long dmax;

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

	dmax = max(rec(N - 2, 0) + A[N - 1], rec(N - 2, 1) - A[N - 1]);

	printf("%ld\n", dmax);
	
	return 0;
}