description: It is the hard version of the problem. The only difference is that in this version 1 ≤ n ≤ 300.
  
  In the cinema seats can be represented as the table with n rows and m columns. The rows are numbered with integers from 1 to n. The seats in each row are numbered with consecutive integers from left to right: in the k-th row from m (k - 1) + 1 to m k for all rows 1 ≤ k ≤ n.
  
  1| 2| ⋅⋅⋅| m - 1| m  
  ---|---|---|---|---  
  m + 1| m + 2| ⋅⋅⋅| 2 m - 1| 2 m  
  2m + 1| 2m + 2| ⋅⋅⋅| 3 m - 1| 3 m  
  \vdots| \vdots| \ddots| \vdots| \vdots  
  m (n - 1) + 1| m (n - 1) + 2| ⋅⋅⋅| n m - 1| n m  
  The table with seats indices
  
  There are nm people who want to go to the cinema to watch a new film. They are numbered with integers from 1 to nm. You should give exactly one seat to each person.
  
  It is known, that in this cinema as lower seat index you have as better you can see everything happening on the screen. i-th person has the level of sight a_i. Let's define s_i as the seat index, that will be given to i-th person. You want to give better places for people with lower sight levels, so for any two people i, j such that a_i < a_j it should be satisfied that s_i < s_j.
  
  After you will give seats to all people they will start coming to their seats. In the order from 1 to nm, each person will enter the hall and sit in their seat. To get to their place, the person will go to their seat's row and start moving from the first seat in this row to theirs from left to right. While moving some places will be free, some will be occupied with people already seated. The inconvenience of the person is equal to the number of occupied seats he or she will go through.
  
  Let's consider an example: m = 5, the person has the seat 4 in the first row, the seats 1, 3, 5 in the first row are already occupied, the seats 2 and 4 are free. The inconvenience of this person will be 2, because he will go through occupied seats 1 and 3.
  
  Find the minimal total inconvenience (the sum of inconveniences of all people), that is possible to have by giving places for all people (all conditions should be satisfied).
  
  Input
  
  The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
  
  The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 300) — the number of rows and places in each row respectively.
  
  The second line of each test case contains n ⋅ m integers a_1, a_2, …, a_{n ⋅ m} (1 ≤ a_i ≤ 10^9), where a_i is the sight level of i-th person.
  
  It's guaranteed that the sum of n ⋅ m over all test cases does not exceed 10^5.
  
  Output
  
  For each test case print a single integer — the minimal total inconvenience that can be achieved.
  
  Example
  
  Input
  
  
  7
  1 2
  1 2
  3 2
  1 1 2 2 3 3
  3 3
  3 4 4 1 1 1 1 1 2
  2 2
  1 1 2 1
  4 2
  50 50 50 50 3 50 50 50
  4 2
  6 6 6 6 2 2 9 6
  2 9
  1 3 3 3 3 3 1 1 3 1 3 1 1 3 3 1 1 3
  
  
  Output
  
  
  1
  0
  4
  0
  0
  0
  1
  
  Note
  
  In the first test case, there is a single way to give seats: the first person sits in the first place and the second person — in the second. The total inconvenience is 1.
  
  In the second test case the optimal seating looks like this: 
  
  <image>
  
  In the third test case the optimal seating looks like this: 
  
  <image>
  
  The number in a cell is the person's index that sits on this place.
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 m = sc.nextInt();
              List<Integer> list = new ArrayList<>();
              for(int i = 0; i < n*m; i++){
                  list.add(sc.nextInt());
              }
              List<Integer> sorted = new ArrayList<>();
              for(int e : list) sorted.add(e);
              Collections.sort(sorted);
              boolean[] vis = new boolean[n*m];
              int[][] arr = new int[n][m];
              for(int[] e : arr) Arrays.fill(e,-1);
              long count = 0;
              for(int i = 0; i < n*m; i++){
                  int val = lowerBound(list.get(i),sorted,vis);
                  int row = val/m;
                  int col = val%m;
                  for(int j = 0; j < col; j++){
                      if(arr[row][j] != -1 && arr[row][j] != list.get(i)){
                          count++;
                      }
                  }
                  arr[row][col] = list.get(i);
                  vis[val] = true;
              }
              out.println(count);
          }
          out.close();
  	}
      
      public static int lowerBound(int curr, List<Integer> sorted, boolean[] vis){
          int start = 0;
          int end = sorted.size()-1;
          int ans = -1;
          while(start <= end){
              int mid = start + (end-start)/2;
              if(sorted.get(mid) == curr){
                  if(vis[mid]){
                      start = mid+1;
                  }
                  else {
                      ans = mid;
                      end = mid-1;
                  }
              }
              else if(sorted.get(mid) > curr){
                  end = mid-1;
              }
              else {
                  start = mid+1;
              }
          }
          return ans;
      }
     
      
      
      /*
      Source: hu_tao
           Random stuff to try when stuck:
              -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) {
  		//for debugging only
  		for (int x : arr)
  			out.print(x + " ");
  		out.println();
  	}
  
  	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);
  }