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 static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.*;
  import java.math.*;
  import java.util.Arrays;
  import java.util.Collections;
  
  
  public class Moamenandksubarrays
  {
  	public static void main(String args[]) throws IOException
  	{
  	    // long start = System.currentTimeMillis();		
  		int u = 0;
  		BufferedReader br = new BufferedReader( new InputStreamReader(System.in));
  		// System.out.println("Enter t");
  		int t = Integer.parseInt(br.readLine());
  		do
  		{
  			// System.out.println("Enter n");
  			int[] input = Arrays.stream(br.readLine().split("\\s+")).mapToInt(Integer::parseInt).toArray();
  			int n = input[0];
  			int k = input[1];
  			long[] input_n = Arrays.stream(br.readLine().split("\\s+")).mapToLong(Long::parseLong).toArray();
  			compress_longarray(input_n);
  			long[] a= new long[n];
  			a = input_n;
  			// System.out.println(input_n.length);
  			// print_longarray(input_n);
  
  			int wall =0;
  			for(int i =0;i<n-1;i++)
  			{
  				if(a[i+1] != a[i] +1)
  				{
  					wall++;
  				}
  				else
  				{
  					continue;
  				}
  
  			}
  			if(wall<k)
  			{
  				System.out.println("YES");
  			}
  			else
  			{
  				System.out.println("NO");
  			}
  
  			
  			
  			
  
  
  
              
              u++;
              
          }
          while(u<t);
          // long end = System.currentTimeMillis();
          // System.out.println("time taken is "+ (end - start) + "ms");
  
  	}
  	
  	public static void compress_longarray(long[] inp)
  	{
  		long[] copy_arr = new long[inp.length];
  
  		copy_arr = inp.clone();
  		Arrays.sort(copy_arr);
  		long val = 0;
  		HashMap<Long,Long> hm = new HashMap<Long,Long>();
  		for(int i =0;i<copy_arr.length;i++)
  		{
  			hm.put(copy_arr[i],val);
  			val++;
  			
  
  		}
  		for(int j=0;j<copy_arr.length;j++)
  		{
  			inp[j] = hm.get(inp[j]);
  		}
  
  	}
  
  }