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 javax.swing.*;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.*;
  
  public class A {
      static ArrayList<Integer> [] adj;
      static ArrayList<Integer> temp;
      static int mod = (int) 1e9+7;
      static boolean[] vis = new boolean[(int)1e5];
      static List<Long> l = new ArrayList<>();
      static int p=0,x=0;
  
      static boolean check(int n){
          for(int i=2;i*i<=n;i++){
              if(n%i==0)return false;
          }
          return true;
      }
      static long lcm(long a, long b) throws IOException {
          return (a / gcd(a, b)) * b;
      }
  
      public static void main(String[] args) throws IOException {
          FastScanner s = new FastScanner();
          int t = s.nextInt();
  //        int t = 1;
          while (t-->0) {
              int n = s.nextInt();
              int k = s.nextInt();
              long[] a =new long[n];
              long[] b =new long[n];
              Map<Long,Integer> map = new HashMap<>();
              for(int i=0;i<n;i++) {
                  a[i] = s.nextInt();
                  b[i]=a[i];
                  map.put(a[i],i);
              }
              sort(b);
              int cnt=0;
              for(int i=1;i<n;i++){
                  if(map.get(b[i])-map.get(b[i-1])!=1)cnt++;
              }
              if(cnt+1<=k) System.out.println("Yes");
              else System.out.println("No");
          }
      }
      //Dont't stick on same approach
      //Check long
  
      static int lower_bound(int[] arr, int x) {
          int low_limit = 0, high_limit = arr.length, mid = -1;
          while (low_limit < high_limit) {
              mid = (low_limit + high_limit) / 2;
              if (arr[mid] >= x) {
                  high_limit = mid;
              } else  {
                  low_limit = mid + 1;
              }
          }
          return low_limit + 1;
      }
  
  
      static void dfs(int i){
          vis[i]=true;
          temp.add(i);
          List<Integer> Y = adj[i];
          for(int x:Y){
              if(!vis[x])dfs(x);
          }
      }
  
      static long gcd(long a, long b) throws IOException {
          return (b == 0) ? a : gcd(b, a % b);
      }
  
      static int gcd(int a, int b) throws IOException {
          return (b == 0) ? a : gcd(b, a % b);
      }
  
      static void sortr(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i : a)
              l.add(i);
          Collections.sort(l, Collections.reverseOrder());
          for (int i = 0; i < a.length; i++)
              a[i] = l.get(i);
      }
  
      static void sort(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i : a)
              l.add(i);
          Collections.sort(l);
          for (int i = 0; i < a.length; i++)
              a[i] = l.get(i);
      }
  
      static void sort(long[] a) {
          ArrayList<Long> l = new ArrayList<>();
          for (long i : a)
              l.add(i);
          Collections.sort(l);
          for (int i = 0; i < a.length; i++)
              a[i] = l.get(i);
      }
  
      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());
          }
      }
  }