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 abs(x)  ((x)>=0 ? (x) : ((x)*-1))		//絶対値


int main(){

    int N, K, i, j;
    scanf("%d", &N);

    long long int B[N], ans = 0;
    for (i=0; i<N; i++){
        scanf("%lld", &B[i]);
    }

    long long int min=10000000000, min_sequence;
    for (i=0; i<N; i++){
        if (abs(B[i]) < min){
            min = abs(B[i]);
            min_sequence = i;
        }
    }

    if (N%2 == 0){
        for (i=0; i<N; i++){
                ans += abs(B[i]);
        }
    }else{
        for (i=0; i<N; i++){
                ans += abs(B[i]);
        }
        ans -= B[min_sequence];
    }
 

    printf("%lld", ans);

    return 0;
}