description: A total of n depots are located on a number line. Depot i lies at the point x_i for 1 ≤ i ≤ n.
  
  You are a salesman with n bags of goods, attempting to deliver one bag to each of the n depots. You and the n bags are initially at the origin 0. You can carry up to k bags at a time. You must collect the required number of goods from the origin, deliver them to the respective depots, and then return to the origin to collect your next batch of goods.
  
  Calculate the minimum distance you need to cover to deliver all the bags of goods to the depots. You do not have to return to the origin after you have delivered all the bags.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10 500). Description of the test cases follows.
  
  The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5).
  
  The second line of each test case contains n integers x_1, x_2, …, x_n (-10^9 ≤ x_i ≤ 10^9). It is possible that some depots share the same position.
  
  It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case, output a single integer denoting the minimum distance you need to cover to deliver all the bags of goods to the depots. 
  
  Example
  
  Input
  
  
  4
  5 1
  1 2 3 4 5
  9 3
  -5 -10 -15 6 5 8 3 7 4
  5 3
  2 2 3 3 3
  4 2
  1000000000 1000000000 1000000000 1000000000
  
  
  Output
  
  
  25
  41
  7
  3000000000
  
  Note
  
  In the first test case, you can carry only one bag at a time. Thus, the following is a solution sequence that gives a minimum travel distance: 0 → 2 → 0 → 4 → 0 → 3 → 0 → 1 → 0 → 5, where each 0 means you go the origin and grab one bag, and each positive integer means you deliver the bag to a depot at this coordinate, giving a total distance of 25 units. It must be noted that there are other sequences that give the same distance.
  
  In the second test case, you can follow the following sequence, among multiple such sequences, to travel minimum distance: 0 → 6 → 8 → 7 → 0 → 5 → 4 → 3 → 0 → (-5) → (-10) → (-15), with distance 41. It can be shown that 41 is the optimal distance for this test case.
solution: //package currentContest;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayDeque;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.HashMap;
  import java.util.Random;
  import java.util.StringTokenizer;
  
  
  public class P1 {
  	
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		FastReader sc=new FastReader();
  		int t=sc.nextInt();
  		StringBuilder st=new StringBuilder();
  		while(t--!=0){
  			int n=sc.nextInt();
  			int k=sc.nextInt();
  			int a[]=new int[n];
  			for(int i=0;i<n;i++) {
  				a[i]=sc.nextInt();
  			}
  			sort(a);
  			ArrayDeque<Integer> tryp=new ArrayDeque<>();
  			ArrayDeque<Integer> trym=new ArrayDeque<>();
  			for(int x:a) {
  				if(x>0)
  					tryp.addFirst(x);
  				else if(x<0) {
  					trym.addLast(x);
  				}
  			}
  			long ans=0;
  			int first=0;
  			int second =0;
  			if(trym.size()>0) {
  				first=Math.abs(trym.removeFirst());
  				int chance=k-1;
  				while(trym.size()>0&&chance>0) {
  					chance--;
  					trym.removeFirst();
  				}
  			}
  			if(tryp.size()>0) {
  				second=Math.abs(tryp.removeFirst());
  				int chance=k-1;
  				while(tryp.size()>0&&chance>0) {
  					chance--;
  					tryp.removeFirst();
  				}
  			}
  			while(tryp.size()>0) {
  				ans+=2*tryp.removeFirst();
  				int chance=k-1;
  				while(tryp.size()>0&&chance>0) {
  					chance--;
  					tryp.removeFirst();
  				}
  			}
  			while(trym.size()>0) {
  				ans+=(long)2*Math.abs(trym.removeFirst());
  				int chance=k-1;
  				while(trym.size()>0&&chance>0) {
  					chance--;
  					trym.removeFirst();
  				}
  			}
  			ans+=(long)2*Math.min(first, second)+(long)Math.max(first, second);
  			st.append(ans+"\n");
  		}
  		System.out.println(st);
  	}
  	
  	static class DisjointUnionSets {
  		int[] rank, parent;
  		int n;
  		
  		// Constructor
  		public DisjointUnionSets(int n)
  		{
  			rank = new int[n];
  			parent = new int[n];
  			this.n = n;
  			makeSet();
  		}
  		
  		// Creates n sets with single item in each
  		void makeSet()
  		{
  			for (int i = 0; i < n; i++) {
  				// Initially, all elements are in
  				// their own set.
  				parent[i] = i;
  			}
  		}
  		
  		// Returns representative of x's set
  		int find(int x)
  		{
  			// Finds the representative of the set
  			// that x is an element of
  			if (parent[x] != x) {
  				// if x is not the parent of itself
  				// Then x is not the representative of
  				// his set,
  				parent[x] = find(parent[x]);
  				
  				// so we recursively call Find on its parent
  				// and move i's node directly under the
  				// representative of this set
  			}
  			
  			return parent[x];
  		}
  		
  		// Unites the set that includes x and the set
  		// that includes x
  		void union(int x, int y)
  		{
  			// Find representatives of two sets
  			int xRoot = find(x), yRoot = find(y);
  			
  			// Elements are in the same set, no need
  			// to unite anything.
  			if (xRoot == yRoot)
  				return;
  			
  			// If x's rank is less than y's rank
  			if (rank[xRoot] < rank[yRoot])
  				
  				// Then move x under y  so that depth
  				// of tree remains less
  				parent[xRoot] = yRoot;
  			
  			// Else if y's rank is less than x's rank
  			else if (rank[yRoot] < rank[xRoot])
  				
  				// Then move y under x so that depth of
  				// tree remains less
  				parent[yRoot] = xRoot;
  			
  			else // if ranks are the same
  			{
  				// Then move y under x (doesn't matter
  				// which one goes where)
  				parent[yRoot] = xRoot;
  				
  				// And increment the result tree's
  				// rank by 1
  				rank[xRoot] = rank[xRoot] + 1;
  			}
  		}
  	}
  
  	static FastReader sc = new FastReader();
  
  	public static void solvegraph() {
  		int n = sc.nextInt();
  
  		int edge[][] = new int[n - 1][2];
  		for (int i = 0; i < n - 1; i++) {
  			edge[i][0] = sc.nextInt() - 1;
  			edge[i][1] = sc.nextInt() - 1;
  		}
  		ArrayList<ArrayList<Integer>> ad = new ArrayList<>();
  		for (int i = 0; i < n; i++) {
  			ad.add(new ArrayList<Integer>());
  		}
  		for (int i = 0; i < n - 1; i++) {
  			ad.get(edge[i][0]).add(edge[i][1]);
  			ad.get(edge[i][1]).add(edge[i][0]);
  		}
  		int parent[] = new int[n];
  		Arrays.fill(parent, -1);
  		parent[0] = n;
  		ArrayDeque<Integer> queue = new ArrayDeque<>();
  		queue.add(0);
  		int child[] = new int[n];
  		Arrays.fill(child, 0);
  		ArrayList<Integer> lv = new ArrayList<Integer>();
  		while (!queue.isEmpty()) {
  			int toget = queue.getFirst();
  			queue.removeFirst();
  			child[toget] = ad.get(toget).size() - 1;
  			for (int i = 0; i < ad.get(toget).size(); i++) {
  				if (parent[ad.get(toget).get(i)] == -1) {
  					parent[ad.get(toget).get(i)] = toget;
  					queue.addLast(ad.get(toget).get(i));
  				}
  			}
  			lv.add(toget);
  		}
  
  		child[0]++;
  	}
  
  	static void sort(int[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			int tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static void sort(long[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			long tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static String sort(String s) {
  		Character ch[] = new Character[s.length()];
  		for (int i = 0; i < s.length(); i++) {
  			ch[i] = s.charAt(i);
  		}
  		Arrays.sort(ch);
  		StringBuffer st = new StringBuffer("");
  		for (int i = 0; i < s.length(); i++) {
  			st.append(ch[i]);
  		}
  		return st.toString();
  	}
  
  	public static long gcd(long a, long b) {
  		if (a == 0) {
  			return b;
  		}
  		return gcd(b % a, a);
  	}
  
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		String next() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  }
  