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.*;
  import java.util.*;
  
  public class Main {
  
      public static void main(String[] args) throws IOException {
  
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  
              int p= Integer.parseInt(br.readLine());
  
              for(int t=0;t<p;t++){
  
                  String s = br.readLine();
                  String[] s1 = s.split(" ");
  
                  int n=Integer.parseInt(s1[0]);
                  int k=Integer.parseInt(s1[1]);
  
                  s = br.readLine();
                  s1 = s.split(" ");
                  int[] arr = new int[n];
                  int[] srr = new int[n];
  
                  for(int i=0;i<n;i++) {
                      arr[i]=Integer.parseInt(s1[i]);
                      srr[i]=Integer.parseInt(s1[i]);
                  }
  
                  Arrays.sort(srr);
  
                  HashMap<Integer, Integer> left = new HashMap<>();
                  HashMap<Integer, Integer> right = new HashMap<>();
  
                  for(int i=0;i<n;i++){
  
                      if(i!=0)
                      left.put(srr[i], srr[i-1]);
  
                      if(i!=n-1)
                      right.put(srr[i], srr[i+1]);
                  }
  
                  int count = n;
  
                  for(int i=0;i<n;i++){
  
                      if(i>0){
                          if(left.get(arr[i])!=null)
                              if(left.get(arr[i]) == arr[i-1]){
                                  count--;
  
                                  right.remove(arr[i-1]);
                              }
                      }
  
  
                      if(i<n-1){
                          if(right.get(arr[i]) != null)
                              if(right.get(arr[i]) == arr[i+1]){
                                  count--;
  
                                  left.remove(arr[i+1]);
                              }
                      }
  
                  }
  
                  if(count<=k)
                      System.out.println("YES");
                  else
                      System.out.println("NO");
              }
  
  
      }
  }
  