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.*;
  
  /**
   * -----------------|___________|---------------------
   * CCCCCCCCC   OOOOOOOOOO   DDDDDDDD      EEEEEEEEE
   * CCCC        OOO    OOO   DD    DDD     EEEE
   * CCCC        OOO    OOO   DD     DDD    EEEEEEEEEE
   * CCCC        OOO    OOO   DD     DDDD   EEEE
   * CCCCCCCCC   OOOOOOOOOO   DDDDDDDDDDDD  EEEEEEEEEE
   * -----------------|___________|---------------------
   */
  
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          InputReader in = new InputReader(inputStream);
          PrintWriter sout = new PrintWriter(outputStream);
          int t = in.nextInt();
          while (t > 0) {
              t--;
              int n = in.nextInt();
              int k = in.nextInt();
              int[] arr = new int[n];
              Map<Integer,Integer> map = new HashMap<>();
              for (int i = 0; i < n; i++) {
                  int val = in.nextInt();
                  arr[i] = val;
                  map.put(val,i);
              }
              int mxk = 1;
              Arrays.sort(arr);
              for(int i = 0; i<n-1; i++){
                  if(map.get(arr[i])+1!=map.get(arr[i+1])){
                      mxk++;
                  }
              }
              sout.println(mxk <= k ? "Yes" : "No");
          }
          sout.close();
      }
  
      public static class InputReader {
          private BufferedReader reader;
          private StringTokenizer tokenizer;
  
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 32768);
              tokenizer = null;
          }
  
          public String nextToken() {
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(nextToken());
          }
  
          public long nextLong() {
              return Long.parseLong(nextToken());
          }
  
          public double nextDouble() {
              return Double.parseDouble(nextToken());
          }
      }
  }
  