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.*;
  
  public class B {
  	static class InputReader {
  		BufferedReader reader;
  		StringTokenizer tokenizer;
  		public InputReader(InputStream stream) {
  			reader = new BufferedReader(new InputStreamReader(stream), 32768);
  			tokenizer = null;
  		}
  		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 ni() {
  			return Integer.parseInt(next());
  		}
  		public long nl() {
  			return Long.parseLong(next());
  		}
  		public double nd() {
  			return Double.parseDouble(next());
  		}
  		
  		public int[] ra(int n ) {
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++) {
  				a[i] = r.ni();
  			}
  			return a;
  		}
  	}
  	
  	public static void sort(int[] a) {
  		List<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 InputReader r = new InputReader(System.in);
  	static PrintWriter pw = new PrintWriter(System.out);
  	public static void main(String[] args) {
  		int TT = r.ni();
  		while (TT --> 0) {
  			int n = r.ni(), k = r.ni();
  			int[] a = new int[n];
  			int[] b = new int[n];
  			for (int i = 0; i < n; i++) {
  				int t = r.ni();
  				a[i] = t;
  				b[i] = t;
  			}
  			sort(b);
  			int sg = 0;
  			
  			for (int i = 0; i < n - 1; i++) {
  				if (a[i] > a[i + 1]) {
  					sg++;
  					
  				} else {
  					int ind = Arrays.binarySearch(b, a[i]);
  					int inds = Arrays.binarySearch(b, a[i+1]);
  					if (inds - ind != 1) {
  						sg++;
  					}
  				}
  				if (i == n - 2) {
  					sg++;
  				}
  				
  				if (sg > k ) {
  					pw.println("NO");
  					break;
  				}
  			}
  			if (k - sg >= 0) {
  				pw.println("YES");
  			}
  		}
  		pw.close();
  	}
  }
  