description: Moamen has an array of n distinct integers. He wants to sort that array in non-decreasing order by doing the following operations in order exactly once:
  
    1. Split the array into exactly k non-empty subarrays such that each element belongs to exactly one subarray. 
    2. Reorder these subarrays arbitrary. 
    3. Merge the subarrays in their new order. 
  
  
  
  A sequence a is a subarray of a sequence b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
  
  Can you tell Moamen if there is a way to sort the array in non-decreasing order using the operations written above?
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^3) — the number of test cases. The description of the test cases follows.
  
  The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 10^5).
  
  The second line contains n integers a_1, a_2, …, a_n (0 ≤ |a_i| ≤ 10^9). It is guaranteed that all numbers are distinct.
  
  It is guaranteed that the sum of n over all test cases does not exceed 3⋅10^5.
  
  Output
  
  For each test case, you should output a single string.
  
  If Moamen can sort the array in non-decreasing order, output "YES" (without quotes). Otherwise, output "NO" (without quotes).
  
  You can print each letter of "YES" and "NO" in any case (upper or lower).
  
  Example
  
  Input
  
  
  3
  5 4
  6 3 4 2 1
  4 2
  1 -4 0 -2
  5 1
  1 2 3 4 5
  
  
  Output
  
  
  Yes
  No
  Yes
  
  Note
  
  In the first test case, a = [6, 3, 4, 2, 1], and k = 4, so we can do the operations as follows: 
  
    1. Split a into \{ [6], [3, 4], [2], [1] \}. 
    2. Reorder them: \{ [1], [2], [3,4], [6] \}. 
    3. Merge them: [1, 2, 3, 4, 6], so now the array is sorted. 
  
  
  
  In the second test case, there is no way to sort the array by splitting it into only 2 subarrays.
  
  As an example, if we split it into \{ [1, -4], [0, -2] \}, we can reorder them into \{ [1, -4], [0, -2] \} or \{ [0, -2], [1, -4] \}. However, after merging the subarrays, it is impossible to get a sorted array.
solution: import java.io.BufferedReader;
  import java.util.StringTokenizer;
  import java.io.InputStreamReader;
  import java.io.IOException;
  import java.io.PrintWriter;
  import java.util.*;
  public class B {
  
  
      public static void main(String[] args) {
          FastReader fr = new FastReader();
          PrintWriter out = new PrintWriter(System.out, true);
          int cases = fr.nextInt();
          for(int c = 0; c < cases; c++) {
              int n = fr.nextInt();
              int k = fr.nextInt();
              int[] nums = new int[n];
              for(int i = 0; i < n; i++) {
                  nums[i] = fr.nextInt();
              }
  
              int inversions = 0;
              int sorted[] = sorted(nums);
              HashMap<Integer, Integer> numlocs = new HashMap<>();
              for(int i = 0; i < sorted.length; i++) {
                  numlocs.put(sorted[i], i);
              }
  
              for(int i = 1; i < n; i++) {
                  if(nums[i] < nums[i-1] || numlocs.get(nums[i]) != (numlocs.get(nums[i-1]) + 1)) {
                      inversions++;
                  }
              }
  
              if(inversions + 1 > k) {
                  out.write("NO\n");
              } else {
                  out.write("YES\n");
              }
  
              
          }
          out.close();
      }
  
      static int[] sorted(int[] nums) {
          ArrayList<Integer> newnums = new ArrayList<>();
          for(int i = 0 ; i < nums.length; i++) {newnums.add(nums[i]);}
          Collections.sort(newnums);
          int ret[] = new int[nums.length];
          for(int i = 0 ;i < nums.length; i++) {ret[i] = newnums.get(i);}
          return ret;
      }
  
      static class FastReader {
  
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              this.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;
          }
      }
  }
  