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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import static java.lang.Math.*;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.File;
  import java.io.PrintStream;
  import java.io.PrintWriter;
  import java.math.BigInteger;
  public class Main {
   
  	
  	/* 10^(7) = 1s.
  	 * ceilVal = (a+b-1) / b */
  	
  	static final int mod = 1000000007;
  	static final long temp = 998244353;
  	static final long MOD = 1000000007;
  	static final long M = (long)1e9+7;
   
  	static class Pair implements Comparable<Pair> 
  	{
  		int first, second;
  		
  		public Pair(int first, int second) 
  		{
  			this.first = first;
  			this.second = second;
  		}
  		
  		public int compareTo(Pair ob) 
  		{
  			return (int)(first - ob.first);
  		}
  	}
  	
  	
  	static class Tuple implements Comparable<Tuple>
  	{
  		int first, second,third;
  		public Tuple(int first, int second, int third)
  		{
  			this.first = first;
  			this.second = second;
  			this.third = third;
  		}
  		public int compareTo(Tuple o)
  		{
  			return (int)(o.third - this.third);
  		}
  	}
  	
  	public static class DSU
  	{
  		int[] parent;
  		int[] rank; //Size of the trees is used as the rank
  		public DSU(int n)
  		{
  			parent = new int[n];
  			rank = new int[n];
  			Arrays.fill(parent, -1);
  			Arrays.fill(rank, 1);
  		}
  		
  		public int find(int i) //finding through path compression
  		{
  			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
  		}
  		
  		public boolean union(int a, int b) //Union Find by Rank 
  		{
  			a = find(a);
  			b = find(b);
  			
  			if(a == b) return false; //if they are already connected we exit by returning false.
  			
  			// if a's parent is less than b's parent
  			if(rank[a] < rank[b])
  			{
  				//then move a under b
  				parent[a] = b;
  			}
  			//else if rank of j's parent is less than i's parent
  			else if(rank[a] > rank[b])
  			{
  				//then move b under a
  				parent[b] = a;
  			}
  			//if both have the same rank.
  			else
  			{
  				//move a under b (it doesnt matter if its the other way around.
  				parent[b] = a;
  				rank[a] = 1 + rank[a];
  			}
  			return true;
  		}	
  	}
  
  	static class Reader 
  	{
  		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) throws IOException {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		
  		long[] longReadArray(int n) throws IOException {
  			long[] a=new long[n];
  			for (int i=0; i<n; i++) a[i]=nextLong();
  			return a;
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  	
  	public static int gcd(int a, int b)
  	{
  		if(b == 0)
  		 return a;
  		
  		else
  		return gcd(b,a%b);
  	} 
   
  		public static long lcm(long a, long b)
  	    {
  	        return (a / LongGCD(a, b)) * b;
  	    }
  	
  	public static long LongGCD(long a, long b)
  	{
  		if(b == 0)
  			 return a;
  			
  			else
  			return LongGCD(b,a%b);
  	}
  	
  		public static long LongLCM(long a, long b)
  	    {
  	        return (a / LongGCD(a, b)) * b;
  	    }
  	
  		//Count the number of coprime's upto N
  		public static long phi(long n)  //euler totient/phi function
  	    { 
  	       long ans = n;
  //	       for(long i = 2;i*i<=n;i++)
  //	       {
  //	    	   if(n%i == 0)
  //	    	   {
  //	    		   while(n%i == 0) n/=i;
  //	    		   ans -= (ans/i);
  //	    	   }
  //	       }
  //	       
  //	       if(n > 1)
  //	       {
  //	    	   ans -= (ans/n);
  //	       }
  	       
  	       for(long i = 2;i<=n;i++)
  	       {
  	    	   if(isPrime(i))
  	    	   {
  	    		   ans -= (ans/i);
  	    	   }
  	       }
  	       
  	       return ans;
  	       
  	    }	
  	
  	public static long fastPow(long x, long n) 
  	{
  		if(n == 0)
  			return 1;
  		else if(n%2 == 0)
  			return fastPow(x*x,n/2);
  		else
  			return x*fastPow(x*x,(n-1)/2);
  	}
  	
  	   public static long modPow(long x, long y, long p)
  	    {
  	        long res = 1;
  	        x = x % p;
  	 
  	        while (y > 0) {
  	            if (y % 2 == 1)
  	                res = (res * x) % p;
  	 
  	            y = y >> 1; 
  	            x = (x * x) % p;
  	        }
  	        return res;
  	    }
  	   
  	   static long modInverse(long n, long p)
  	    {
  	        return modPow(n, p - 2, p);
  	    }
  	 
  	    // Returns nCr % p using Fermat's little theorem.
  	    
  	    public static long nCrModP(long n, long r,long p)
  	    {
  	    	if (n<r) 
  	              return 0;
  	          
  	        if (r == 0)
  	            return 1;
  	 
  	        long[] fac = new long[(int)(n) + 1];
  	        fac[0] = 1;
  	 
  	        for (int i = 1; i <= n; i++)
  	            fac[i] = fac[i - 1] * i % p;
  	 
  	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
  	                % p * modInverse(fac[(int)(n - r)], p)
  	                % p)
  	            % p;
  	    }
  	    
  	    public static long fact(long n) 
  	    {
  	    	  long[] fac = new long[(int)(n) + 1];
  		        fac[0] = 1;
  		 
  		        for (long i = 1; i <= n; i++)
  		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
  		        
  		        return fac[(int)(n)];
  		}
  	    
  	    public static long nCr(long n, long k) 
  	    {
  			long ans = 1;
  			for(long i = 0;i<k;i++)
  			{
  				ans *= (n-i);
  				ans /= (i+1);
  			}
  			return ans;
  		}
   
  		//Modular Operations for Addition and Multiplication. 
  		   public static long perfomMod(long x)
  		   {
  		        return ((x%M + M)%M);
  		    }
  		   public static long addMod(long a, long b)
  		   {
  		        return perfomMod(perfomMod(a)+perfomMod(b));
  		    }
  		  
  		   public static long subMod(long a, long b)
  		   {
  		        return perfomMod(perfomMod(a)-perfomMod(b));
  		   }
  		   
  		   public static long mulMod(long  a, long b)
  		   {
  		        return perfomMod(perfomMod(a)*perfomMod(b));
  		   }
  		   
  		   public static boolean isPrime(long n) 
  			{
  				if(n == 1)
  				{
  					return false;
  				}
  				for(int i = 2;i*i<=n;i++)
  				{
  					if(n%i == 0)
  					{
  						return false;
  					}
  				}
  				return true;
  			}
  			
  			public static List<Long> SieveList(int n)
  			{
  				boolean prime[] = new boolean[(int)(n+1)];
  				Arrays.fill(prime, true);
  				
  				List<Long> l = new ArrayList<>();
  				for (long p = 2; p*p<=n; p++) 
  				{ 
  					if (prime[(int)(p)] == true) 
  					{ 	
  					    for(long i = p*p; i<=n; i += p) 
  						{
  						    prime[(int)(i)] = false; 
  						}	
  					} 
  				} 
  		 
  				for (long p = 2; p<=n; p++) 
  				{
  				    if (prime[(int)(p)] == true)
  				    {
  				       l.add(p); 
  				    }
  				}
  				
  				return l;
  			}
  
  			public static long log2(long n)
  		    {
  				 long ans = (long)(log(n)/log(2));
  				 return ans;
  			}
  			   
  			public static boolean isPow2(long n)
  			{
  				 return (n  != 0 && ((n & (n-1))) == 0);
  			}	
  		   
  		   public static boolean isSq(int x)  
  		   {
  			    long s = (long)Math.round(Math.sqrt(x));
  			    return s*s==x;
  		   }
  		
  			   /*
  			    * 
  			    * >=         <=
  			   0 1 2 3 4 5 6 7  
  			   5 5 5 6 6 6 7 7
  			   lower_bound for 6 at index 3 (>=)
  			   upper_bound for 6 at index 6(To get six reduce by one) (<=)
  			   */
  
  		   public static int LowerBound(long a[], long x) 
  		   {
  			      int l=-1,r=a.length;
  				  while(l+1<r) 
  				  {
  				    int m=(l+r)>>>1;
  				    if(a[m]>=x) r=m;
  				    else l=m;
  				  }
  				  return r;
  			}
  		   
  			public static int UpperBound(long a[], long x) 
  			{
  				int l=-1, r=a.length;
  			    while(l+1<r) 
  			    {
  			       int m=(l+r)>>>1;
  			       if(a[m]<=x) l=m;
  			       else r=m;
  			    }
  			    return l+1;
  			}
  			
  			public static void Sort(long[] a) 
  			{
  				List<Long> l = new ArrayList<>();
  				for (long i : a) l.add(i);
  				Collections.sort(l);
  //				Collections.reverse(l);  //Use to Sort decreasingly 
  				for (int i=0; i<a.length; i++) a[i]=l.get(i);
  			}
  			
  			public static void ssort(char[] a)
  			{
  				List<Character> l = new ArrayList<>();
  				for (char i : a) l.add(i);
  				Collections.sort(l);
  				for (int i=0; i<a.length; i++) a[i]=l.get(i);
  			}
  	
  
  	
  	public static void main(String[] args) throws Exception 
  	{
  		Reader sc = new Reader();
  		PrintWriter fout = new PrintWriter(System.out);
  		
  		int t = sc.nextInt();
  		while(t-- > 0)
  		{
  			int n = sc.nextInt(), k = sc.nextInt();
  			long[] a = sc.longReadArray(n);
  			
  			long[] b = new long[n];
  			for(int i = 0;i<n;i++) b[i] = a[i];
  			
  			Sort(b);
  			HashMap<Long,Integer> map = new HashMap<>();
  			for(int i = 0;i<n;i++)
  			{
  				map.put(b[i],i);
  			}
  			
  			long c = 1;
  			for(int i = 1;i<n;i++)
  			{
  				long x = map.get(a[i]);
  				long y = map.get(a[i-1])+1;
  				if(x != y) c++;
  			}
  			
  			fout.println((c <= k) ? "Yes" : "No");
  		}
  
  		fout.close();
  	}
  }