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: //package godiji;
  
  import java.util.Scanner;
  import java.util.Stack;
  import java.util.StringTokenizer;
  import java.util.TreeMap;
  
  import java.io.BufferedReader;
  import java.io.BufferedWriter;
  import java.io.FileWriter;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.math.BigInteger;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.Comparator;
  import java.util.HashMap;
  import java.util.HashSet;
  import java.util.Iterator;
  import java.util.LinkedList;
  import java.util.PriorityQueue;
  
  public class pac1 {
  	static Scanner scn = new Scanner(System.in);
  	static long mod = 1000000007;
  
  	public static void main(String[] args) {
  		int t = scn.nextInt();
  		while (t-- > 0) {
  			int n = scn.nextInt(), k = scn.nextInt();
  			int arr[] = new int[n], brr[] = new int[n];
  			HashMap<Integer, Integer>map=new HashMap<Integer, Integer>();
  			//TreeMap<Integer,Integer>srtmap=new TreeMap<Integer, Integer>();
  			for (int i = 0; i < n; i++) {
  				arr[i] = scn.nextInt();
  				map.put(arr[i],i);
  				brr[i] = arr[i];
  			}
  			Arrays.sort(brr);
  			
  			//System.out.println(map+" "+srtmap);
  			if (n == 1)
  				System.out.println("YES");
  			else 
  			{
  				int count=0,i=0,j=1;
  				while(j<n)
  				{
  					if(map.get(brr[j])-map.get(brr[i])==1)
  					{
  						while(j<n&&map.get(brr[j])-map.get(brr[i])==1) 
  						{
  							i++;
  							j++;
  						}
  						
  					}
  					else
  					{
  						count++;
  						i++;j++;
  					}
  				}
  				
  				//System.out.println(count+1);
  				if(k>=count+1)
  					System.out.println("YES");
  				else
  					System.out.println("NO");
  			}
  
  		}
  
  	}
  
  	public static long coin(int arr[], int i, int j, int count) {
  		if (count == arr.length || i == arr.length) {
  			return 0;
  		}
  		if (i > arr.length || j < 0)
  			return Integer.MIN_VALUE;
  
  		long a = 0, b = 0, c = 0, d = 0;
  		if (count % 2 == 0)
  			a = coin(arr, i + 1, j, count + 1) + arr[i];
  		if (count % 2 == 0)
  			b = coin(arr, i, j - 1, count + 1) + arr[j];
  
  		if (count % 2 != 0)
  			c = coin(arr, i + 1, j, count + 1);
  		if (count % 2 != 0)
  			d = coin(arr, i, j - 1, count + 1);
  
  		return Math.max(a, Math.max(b, Math.max(c, d)));
  
  	}
  
  	public static int gcd(int a, int n) {
  
  		if (a == 0)
  			return n;
  		return gcd(n % a, a);
  	}
  
  	public static boolean[] sieveOfEratosthenes(int n) {
  
  		boolean prime[] = new boolean[n + 1];
  		for (int i = 0; i < n; i++)
  			prime[i] = true;
  
  		for (int p = 2; p * p <= n; p++) {
  			// If prime[p] is not changed, then it is a prime
  			if (prime[p] == true) {
  				// Update all multiples of p
  				for (int i = p * p; i <= n; i += p)
  					prime[i] = false;
  			}
  		}
  
  		for (int i = 2; i <= n; i++) {
  			if (prime[i] == true) {
  				// list.add(i);
  			}
  
  		}
  		return prime;
  	}
  
  }
  
  class pair {
  	int amt;
  	int x, y;
  
  	public pair(int a, int b, int c) {
  		this.amt = a;
  		this.x = b;
  		this.y = c;
  	}
  
  	@Override
  	public boolean equals(Object obj) {
  		// TODO Auto-generated method stub
  		return super.equals(obj);
  	}
  
  	@Override
  	public int hashCode() {
  		// TODO Auto-generated method stub
  		return super.hashCode();
  	}
  }
  