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.*;
  import java.lang.*;
  import java.io.*;
  
  public class A {
  
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  
      int gcd(int a, int b) {
          if (b == 0) return a;
          return gcd(b, a%b);
      }
  
  
      // Map<Integer, Integer> map = new HashMap<>();
      // Set<Integer> set = new HashSet<>();
      // List<Integer> l = new ArrayList<>();
      // StringBuilder sb = new StringBuilder();
  
  
      public static void main (String[] args) throws java.lang.Exception {
  
  //        Reader sc = new Reader();
          FastScanner sc = new FastScanner();
  //        PrintWriter out=new PrintWriter(System.out);
          int tc = sc.nextInt();
  
          l:
          while(tc-- > 0) {
  
  
              /********************************* Code Here ************************************/
              int n = sc.nextInt();
              int m = sc.nextInt();
              int[] a = sc.readArray(n);
              int[] s = new int[n];
              for (int i = 0; i < n; i++) {
                  s[i] = a[i];
              }
  
              Arrays.sort(s);
              int cnt = 0;
  
              for (int i = 1; i < n; i++) {
                  if (a[i-1] > a[i]) {
                      cnt++;
                  } else {
                      int ind = Arrays.binarySearch(s, a[i-1]);
                      if (ind == n-1 || s[ind+1] != a[i]) cnt++;
                  }
              }
  
  //            System.out.println(cnt);
              System.out.println((cnt > m-1) ? "No" : "Yes");
  //            out.close();
          }
      }
  }
  