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.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.util.*;
  import java.io.IOException;
  import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.io.InputStream;
  
  public class java1 {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          InputReader in = new InputReader(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          TaskB solver = new TaskB();
          solver.solve(1, in, out);
          out.close();
      }
      static class TaskB {
  
          public void solve(int testNumber, InputReader in, PrintWriter out) {
              int t=in.nextInt();
              while(t-- >0)
              {
              	int n=in.nextInt();
              	int k=in.nextInt();
              	int ar[]=in.inputarInt(n);
              	int so[]=ar.clone();
              	sort(so);
              	int count=0;
              	int lcount=0;
              	for(int x=0;x<n;x++)
              	{
              		int index=Arrays.binarySearch(so, ar[x]);
              		//out.println(ar[x]+"  "+index);
              		if(x<n && (index-1)<n)
              		{
              			for(int y=(index); y<n-1 && x<n-1;y++)
              			{
              				if(ar[x+1]==so[y+1])
              				{
              					//out.println("runs"+(x+1)+" "+(y+1)+"  "+ar[x+1]+" "+so[y+1]) ;
              					x++;
              				}
              				else
              				{
              					break;
              				}
              			}
              	
              		}
              		count++;
              	}
              	if(count<=k)
              	{
              		out.println("Yes ");
              	}
              	else
              	{
              		out.println("No");
              	}
              	
              	
              	
              			
              }
          }
          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 class InputReader {
          public BufferedReader reader;
          public StringTokenizer tokenizer;
  
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 32768);
              tokenizer = null;
          }
          public String next() {
              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(next());
          }
          public long nextLong() {
              return Long.parseLong(next());
          }
          public int[] inputarInt(int n)
          {
          	int ar[]=new int[n];
          	for(int x=0;x<n;x++)
          	{
          		ar[x]=nextInt();
          	}
          	return ar;
          }
          public long[] inputarLong(int n)
          {
          	long ar[]=new long[n];
          	for(int x=0;x<n;x++)
          	{
          		ar[x]=nextLong();
          	}
          	return ar;
          }
      }
  }