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>
#include <string.h>
#include <stdlib.h>
int main(void)
{
	int n, i, j;
	scanf("%d",&n);
	int a[n];
	for(i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	
	for(i=0;i<n;i++){
		if(a[i]<0){
			if(i+1<n){
				if(a[i+1]<0){
					a[i]=-1*a[i];
					a[i+1]=-1*a[i+1];
					i=i+1;
				}
				if(i+2<n){
					if(a[i+1]>0 && a[i+2]<0){
						a[i]=-1*a[i];
						a[i+2]=-1*a[i+2];
						i=i+2;
					}
				}
			}
		}
	}
	
	long long abs_t = 0;
	int count = 0;
	for(i=0;i<n;i++){
		abs_t += abs(a[i]);
		if(a[i]<0){
			count++;
		}
	}
	int tmp;
	for (i=0;i<n;i++) {
		for (j=i+1;j<n;j++) {
			if (abs(a[i]) > abs(a[j])) {
				tmp =  a[i];
				a[i] = a[j];
				a[j] = tmp;
			}
		}
	}

	for (i=0;i<count;i++) {
		abs_t -= 2*a[i];
	}
	
	printf("%lld",abs_t);
	return 0;
}