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: import static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.*;
  import java.math.*;
  /*
  -> Give your 100%, that's it!
  -> Rules To Solve Any Problem:
      1. Read the problem.
      2. Think About It.
      3. Solve it!
  */
  
  public class Template {
      
  	static int mod = 1000000007;
  
  	public static void main(String[] args){
          FastScanner sc = new FastScanner();
          PrintWriter out = new PrintWriter(System.out);
          int yo = sc.nextInt();
          while (yo-- > 0) {
              int n = sc.nextInt();
              int k = sc.nextInt();
              int[] a = sc.readInts(n);
              sort(a);
              long tot = 0;
              tot += pos(a,n,k);
              tot += neg(a,n,k);
              // out.println(tot);
              if(a[0] < 0 && a[n-1] >= 0){
                  long op1 = tot + 0l + abs(a[0]);
                  long op2 = tot + 0l + a[n-1];
                  tot = Math.min(op1,op2);
              }
              out.println(tot);
          }
          out.close();
  	}
      
      static long neg(int[] a, int n, int k){
          if(a[0] >= 0) return 0l;
          
          int index = n;
          for(int i = 0; i < n; i++){
              if(a[i] >= 0){
                  index = i;
                  break;
              }
          }
          
          long count = 0;
          for(int i = 0; i < index; i += k){
              if(i == 0) count += -a[i];
              else count += 2l * -a[i];
          }
          return count;
          
      }
      
      static long pos(int[] a, int n, int k){
          if(a[n-1] < 0) return 0l;
          int index = 0;
          for(int i = 0; i < n; i++){
              if(a[i] >= 0){
                  index = i;
                  break;
              }
          }
          long count = 0;
          for(int i = n-1; i >= index; i -= k){
              if(i == n-1) count += a[i];
              else count += 2l * a[i];
          }
          return count;
      }
     
      
      
      /*
      Source: hu_tao
           Random stuff to try when stuck:
              - use bruteforcer
              - check for n = 1, n = 2, so on
              -if it's 2C then it's dp
              -for combo/probability problems, expand the given form we're interested in
              -make everything the same then build an answer (constructive, make everything 0 then do something)
              -something appears in parts of 2 --> model as graph
              -assume a greedy then try to show why it works
              -find way to simplify into one variable if multiple exist
              -treat it like fmc (note any passing thoughts/algo that could be used so you can revisit them)
              -find lower and upper bounds on answer
              -figure out what ur trying to find and isolate it
              -see what observations you have and come up with more continuations
              -work backwards (in constructive, go from the goal to the start)
              -turn into prefix/suffix sum argument (often works if problem revolves around adjacent array elements)
              -instead of solving for answer, try solving for complement (ex, find n-(min) instead of max)
              -draw something
              -simulate a process
              -dont implement something unless if ur fairly confident its correct
              -after 3 bad submissions move on to next problem if applicable
              -do something instead of nothing and stay organized
              -write stuff down
           Random stuff to check when wa:
              -if code is way too long/cancer then reassess
              -switched N/M
              -int overflow
              -switched variables
              -wrong MOD
              -hardcoded edge case incorrectly
           Random stuff to check when tle:
              -continue instead of break
              -condition in for/while loop bad
           Random stuff to check when rte:
              -switched N/M
              -long to int/int overflow
              -division by 0
              -edge case for empty list/data structure/N=1
        */
  
  	public static class Pair {
  		int x;
  		int y;
  
  		public Pair(int x, int y) {
  			this.x = x;
  			this.y = y;
  		}
  	}
  
  	public static void sort(int[] arr) {
  		ArrayList<Integer> ls = new ArrayList<Integer>();
  		for (int x : arr)
  			ls.add(x);
  		Collections.sort(ls);
  		for (int i = 0; i < arr.length; i++)
  			arr[i] = ls.get(i);
  	}
  
  	public static long gcd(long a, long b) {
  		if (b == 0)
  			return a;
  		return gcd(b, a % b);
  	}
  
  	static boolean[] sieve(int N) {
  		boolean[] sieve = new boolean[N + 1];
  		for (int i = 2; i <= N; i++)
  			sieve[i] = true;
  
  		for (int i = 2; i <= N; i++) {
  			if (sieve[i]) {
  				for (int j = 2 * i; j <= N; j += i) {
  					sieve[j] = false;
  				}
  			}
  		}
  		return sieve;
  	}
  
  	public static long power(long x, long y, long p) {
  		long res = 1L;
  		x = x % p;
  		while (y > 0) {
  			if ((y & 1) == 1)
  				res = (res * x) % p;
  			y >>= 1;
  			x = (x * x) % p;
  		}
  		return res;
  	}
  
  	public static void print(int[] arr, PrintWriter out) {
  		//for debugging only
  		for (int x : arr)
  			out.print(x + " ");
  		out.println();
  	}
      
      public static int log2(int a){
          return (int)(Math.log(a)/Math.log(2));
      }
      
      public static long ceil(long x, long y){
          return (x + 0l + y - 1) / y;
      }
  
  	static class FastScanner {
  		private int BS = 1 << 16;
  		private char NC = (char) 0;
  		private byte[] buf = new byte[BS];
  		private int bId = 0, size = 0;
  		private char c = NC;
  		private double cnt = 1;
  		private BufferedInputStream in;
  
  		public FastScanner() {
  			in = new BufferedInputStream(System.in, BS);
  		}
  
  		public FastScanner(String s) {
  			try {
  				in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
  			} catch (Exception e) {
  				in = new BufferedInputStream(System.in, BS);
  			}
  		}
  
  		private char getChar() {
  			while (bId == size) {
  				try {
  					size = in.read(buf);
  				} catch (Exception e) {
  					return NC;
  				}
  				if (size == -1)
  					return NC;
  				bId = 0;
  			}
  			return (char) buf[bId++];
  		}
  
  		public int nextInt() {
  			return (int) nextLong();
  		}
  
  		public int[] readInts(int N) {
  			int[] res = new int[N];
  			for (int i = 0; i < N; i++) {
  				res[i] = (int) nextLong();
  			}
  			return res;
  		}
  
  		public long[] readLongs(int N) {
  			long[] res = new long[N];
  			for (int i = 0; i < N; i++) {
  				res[i] = nextLong();
  			}
  			return res;
  		}
  
  		public long nextLong() {
  			cnt = 1;
  			boolean neg = false;
  			if (c == NC)
  				c = getChar();
  			for (; (c < '0' || c > '9'); c = getChar()) {
  				if (c == '-')
  					neg = true;
  			}
  			long res = 0;
  			for (; c >= '0' && c <= '9'; c = getChar()) {
  				res = (res << 3) + (res << 1) + c - '0';
  				cnt *= 10;
  			}
  			return neg ? -res : res;
  		}
  
  		public double nextDouble() {
  			double cur = nextLong();
  			return c != '.' ? cur : cur + nextLong() / cnt;
  		}
  
  		public double[] readDoubles(int N) {
  			double[] res = new double[N];
  			for (int i = 0; i < N; i++) {
  				res[i] = nextDouble();
  			}
  			return res;
  		}
  
  		public String next() {
  			StringBuilder res = new StringBuilder();
  			while (c <= 32)
  				c = getChar();
  			while (c > 32) {
  				res.append(c);
  				c = getChar();
  			}
  			return res.toString();
  		}
  
  		public String nextLine() {
  			StringBuilder res = new StringBuilder();
  			while (c <= 32)
  				c = getChar();
  			while (c != '\n') {
  				res.append(c);
  				c = getChar();
  			}
  			return res.toString();
  		}
  
  		public boolean hasNext() {
  			if (c > 32)
  				return true;
  			while (true) {
  				c = getChar();
  				if (c == NC)
  					return false;
  				else if (c > 32)
  					return true;
  			}
  		}
  	}
  
  	//	For Input.txt and Output.txt	
  	//	FileInputStream in = new FileInputStream("input.txt");
  	//	FileOutputStream out = new FileOutputStream("output.txt");
  	//	PrintWriter pw = new PrintWriter(out);
  	//	Scanner sc = new Scanner(in);
  }