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<string.h>
#include<math.h>

int max(int a,int b){
	if(a>b)return a;
	return b;
}

int min(int a,int b){
	if(a<b)return a;
	return b;
}

int compare_char(char *a,char *b){
	return strcmp(a,b);
}

int intsort(const void *a, const void *b){
	return *(int*)b - *(int*)a;
	/*qsort(a,aの配列数,sizeof(int),intsort);*/
}

int longsort(const void *a, const void *b){
	long *A=(long *)a;
	long *B=(long *)b;
	if(*A>*B) return -1;
	if(*A<*B) return 1;
	return 0;
}

int main(){
  int i,N,t=1;
  long min,cnt,now;
  scanf("%d",&N);
  scanf("%ld",&min);
  if(min<0)t=-1;
  min=labs(min);
  cnt=min;
  for(i=1;i<N;i++){
    scanf("%ld",&now);
    if (now<0){
      t*=-1;
    }
    now=labs(now);
    if(min>now)min=now;
    cnt+=now;
  }
  if(t==1){
    printf("%ld\n",cnt);
  }else{
    printf("%ld\n",cnt-min*2);
  }
  return 0;
}