Roma works in a company that sells TVs. Now he has to prepare a report for the last year.

Roma has got a list of the company's incomes. The list is a sequence that consists of n integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly k changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times.

The operation of changing a number's sign is the operation of multiplying this number by -1.

Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly k changes.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 105), showing, how many numbers are in the sequence and how many swaps are to be made.

The second line contains a non-decreasing sequence, consisting of n integers ai (|ai| ≤ 104).

The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order.

Output

In the single line print the answer to the problem — the maximum total income that we can obtain after exactly k changes.

Sample test(s)

input

3 2-1 -1 1

output

3

input

3 1-1 -1 1

output

1

Note

In the first sample we can get sequence [1, 1, 1], thus the total income equals 3.

In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1.

要注意的是K次操作必须全都执行,所以在tk=n,k的最小值里做循环,若a小于0则乘以-1,tk-1

另:第一次的ia的sort可以省略

import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;import java.util.Arrays;public class CF262B {	public static void main(String[] args) throws IOException {		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));		String[] nk = in.readLine().split(" ");		int n = Integer.parseInt(nk[0]);		int k = Integer.parseInt(nk[1]);		String[] sa = in.readLine().split(" ");		int[] ia = new int[sa.length];		for (int i = 0; i < sa.length; i++) {			ia[i] = Integer.parseInt(sa[i]);		}		//Arrays.sort(ia);		int tk = k;		for (int i = 0; i < Math.min(n, k); i++) {			if (ia[i] < 0) {				ia[i] *= -1;				tk--;			}		}		if (tk > 0 && (tk & 1) == 1) {			Arrays.sort(ia);			ia[0] = ia[0] * -1;		}		int sum = 0;		for (int i = 0; i < n; i++) {			sum += ia[i];		}		System.out.println(sum);	}}