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: #include <bits/stdc++.h>
  using namespace std;
  template <typename A>
  ostream &operator<<(ostream &cout, vector<A> const &v) {
    cout << "[";
    for (int i = 0; i < v.size(); i++) {
      if (i) cout << ", ";
      cout << v[i];
    }
    return cout << "]";
  }
  template <typename A, typename B>
  ostream &operator<<(ostream &cout, pair<A, B> const &p) {
    return cout << "(" << p.first << ", " << p.second << ")";
  }
  template <typename A, typename B>
  istream &operator>>(istream &cin, pair<A, B> &p) {
    cin >> p.first;
    return cin >> p.second;
  }
  int drx[] = {1, -1, 0, 0};
  int dry[] = {0, 0, 1, -1};
  void caseNumber(int x) { printf("Case %d: ", x); }
  void yes() { cout << "YES\n"; }
  void no() { cout << "NO\n"; }
  int gray_code(int n) { return n ^ (n >> 1); }
  int turnOn(int x, int pos) { return x | (1 << pos); }
  int turnOff(int x, int pos) { return x & !(1 << pos); }
  bool isOn(int x, int pos) { return (bool)(x & (1 << pos)); }
  double geo_dis(double x1, double y1, double x2, double y2) {
    double d1 = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
    d1 = sqrt(d1);
    return d1;
  }
  long long int const mod_t = 1000000007;
  long long int const inf = 3e18 + 9;
  long long int const mod = 1e9 + 7;
  long long int const MAX = 1000005;
  long long int const maxn = 1e3 + 9;
  long long int const new_mod = 998244353;
  bool comp(pair<int, int> a, pair<int, int> b) { return a.first < b.first; }
  void solve() {
    int n, k;
    cin >> n >> k;
    vector<pair<int, int> > a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i].first;
      a[i].second = i;
    }
    sort(a.begin(), a.end(), comp);
    int c = 1;
    for (int i = 1; i < n; i++) {
      if (a[i].second - 1 == a[i - 1].second) {
      } else
        c++;
    }
    if (c <= k)
      cout << "Yes" << endl;
    else
      cout << "No" << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int tc;
    cin >> tc;
    while (tc--) solve();
  }
  