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.util.*;
  
  public class Div2_737B {
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int t = sc.nextInt();
          while (t-- > 0) {
              int n = sc.nextInt();
              long k = sc.nextLong();
              long count = 0;
              long arr[] = new long[n];
              for (int i = 0; i < n; i++)
                  arr[i] = sc.nextLong();
              if(n==0) System.out.println("No");
              else if (k == n) {
                  System.out.println("Yes");
              } else {
                  long sortedArr[] = new long[n];
                  for (int i = 0; i < n; i++)
                      sortedArr[i] = arr[i];
                  Arrays.sort(sortedArr);
                  HashMap<Long, Long> map = new HashMap<>();
                  for (int i = 0; i < n - 1; i++) {
                      map.put(sortedArr[i], sortedArr[i + 1]);
                  }
                  map.put(sortedArr[n - 1], Long.MIN_VALUE);
                  boolean flag=false;
                  for (int i = 0; i < n - 1; i++) {
                      flag=false;
                      if (arr[i] > arr[i + 1])
                          count++;
                      else {
                          long next = map.get(arr[i]);
                          if (arr[i + 1] != next){
                              count++;
                              flag=true;
                          }                            
                      }
                  }                
                  if ((flag && (arr[n - 1] == sortedArr[n - 1]))||(count!=0 && !flag) || (arr[n - 1] != sortedArr[n - 1]))
                      count++;
                                  
  
                  //System.out.println(count);
                  if (count <= k)
                      System.out.println("Yes");
                  else
                      System.out.println("No");
              }
          }
      }
  }
  