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 java.util.*;
  import java.util.stream.*;
  
  public class Main {
      public static void main(String[] args) throws IOException{
          MoamenAndKSubarrays ob = new MoamenAndKSubarrays();
          ob.solution(new FastScanner());
      }
  }
  
  class FastScanner {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer token = new StringTokenizer("");
          
          String next() { 
              while (!token.hasMoreTokens()) {
                  try {
                      token = new StringTokenizer(br.readLine());
                  } catch (IOException ie) {
                      ie.printStackTrace();
                  }
              }
              return token.nextToken();
          }
          
          int nextInt () {
              return Integer.parseInt(next());
          }
          long nextLong () {
              return Long.parseLong(next());
          }
          double nextDouble() {
              return Double.parseDouble(next());
          }
          
          int [] readArray (int n) {
              int [] a = new int[n];
              for (int i = 0; i < n; ++i) {
                  a[i] = this.nextInt();
              }
              return a;
          }
      }
  
  class EzzatAndTwoSubsequence {
      public void solution (FastScanner fs) {
          int tt = fs.nextInt();
          for (int qq = 0; qq < tt; ++qq) {
              int n = fs.nextInt();
              int [] arr = fs.readArray(n);
              double ans = 0;
              int maxi = -(int)2e9;
              for (int i = 0; i < n; ++i) {
                  maxi = Math.max(maxi, arr[i]);
                  ans += arr[i];
              }
              ans = (ans - maxi) / (n - 1.0) + maxi;
              System.out.println(ans);
          }
      }
  }
  
  class MoamenAndKSubarrays {
      public void solution(FastScanner fs) {
          int tt = fs.nextInt();
          for (int qq = 0; qq < tt; ++qq) {
              int n = fs.nextInt(), k = fs.nextInt();
              int [] a = fs.readArray(n);
              int [] b = a.clone();
              Arrays.sort(b);
              Map<Integer, Integer> next = new HashMap<Integer, Integer> (); 
              for (int i = 0; i < n - 1; ++i) {
                  if(b[i] != b[i + 1])
                      next.put(b[i], b[i + 1]);
              }
              int partitions = 0;
              for (int i = 0; i < n - 1; ++i) {
                  Integer value = next.get(a[i]);
                  if (value == null) 
                      value = -(int)2e9;
                  if (value != a[i +1])
                      partitions++;
              }
              if (partitions + 1 <= k) 
                  System.out.println("YES");
              else 
                  System.out.println("NO");
          }
      }
  }
  
  
  
  class MoamenAndXor {
      final int MOD = (int)1e9 + 7;
      public void solution(FastScanner fs) {
          int tt = fs.nextInt();
          for (int qq = 0; qq < tt; ++qq) {
              int n = fs.nextInt(), k = fs.nextInt();
              int ans = 0;
              if(k == 0) {
                  ans = 1;
              } else if ((n & 1) == 0) { // even 
                  k = log2(k);
                  System.out.println("k = " + k);
                  for (int i = 1; i <= k; ++i) {
                      int right = modPow(2, modMul(k - i, n));
                      int left = modPow(2, modMul(i - 1, n - 1));
                      ans = modAdd(ans, modMul(right, left));
                  }
                  System.out.println("ans = " + ans);
                  ans = modAdd(ans, modAdd(modPow(2, modMul(n, k - 1)), 1));
  
              } else { // odd
                  ans = modPow(modAdd(modPow(2, n - 1), 1), k);
                  for (int i = 2; i < n; i += 2) {
                      ans = modSub(ans, modMul(nCk(n, i), modSub(modPow(2, modMul(k - 1, i)), 1)));
                  }
                  ans = modSub(ans, modMul(nCk(n, n), modSub(modPow(2, modMul(k - 1, n)), 1)));
              }
              System.out.println(ans);
          }
      }
  
      public static int log2(int n) {
          return (int)((Math.log(n) / Math.log(2)));
      }
  
      // stufff to do
      private int nCk(int n, int k) {
          return 0;
      }
  
      private int modAdd (int a, int b) {
          return ((a + b) % MOD + MOD) % MOD;
      }
      private int modMul (int a, int b) {
          return (int) (((1L * a * b) % MOD + MOD) % MOD);
      }
      private int modSub (int a, int b) {
          return ((a - b) % MOD + MOD) % MOD;
      }
      private int modPow (int a, int b ) {
          int res = 1;
          for (; b > 0; b >>= 1) {
              if ((b & 1) > 0)
                  res = modMul(res, a);
              a = modMul(a, a); 
          }
          return res;
      }
  }