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;
  double Ans = 0;
  long long mod = 100000000;
  struct heap1 {
    int n;
    int i;
    vector<int> v;
    heap1(int n_) {
      n = n_;
      i = 0;
      v.resize(n_);
    }
    void insert(int x) { v[i++] = x; }
    int remove_min() {
      if (i == 0) {
        return -1;
      }
      int j = 0;
      for (int k = 0; k < i; ++k) {
        if (v[k] < v[j]) {
          j = k;
        }
      }
      swap(v[j], v[i - 1]);
      return v[--i];
    }
    ~heap1() {}
  };
  struct heap2 {
    int n;
    int i;
    vector<int> v;
    heap2(int n_) {
      n = n_;
      i = 0;
      v.resize(n_);
    }
    void insert(int x) {
      int j = i;
      v[i++] = x;
      while (j > 0 && v[j - 1] < v[j]) {
        swap(v[j - 1], v[j]);
        j--;
      }
    }
    int remove_min() {
      if (i == 0) {
        return -1;
      }
      return v[--i];
    }
  };
  struct bin_heap {
    int i;
    vector<int> v;
    bin_heap(int n) {
      i = 0;
      v.resize(n);
    }
    void insert(int x) {
      int j = i;
      v[i++] = x;
      while (j > 0 && v[(j - 1) / 2] > v[j]) {
        swap(v[(j - 1) / 2], v[j]);
        j = (j - 1) / 2;
      }
    }
    int remove_min() {
      if (i == 0) {
        return -1;
      }
      int res = v[0];
      v[0] = v[--i];
      int j = 0;
      int m;
      while (true) {
        m = j;
        if (2 * j + 1 < i && v[2 * j + 1] < v[m]) {
          m = 2 * j + 1;
        }
        if (2 * j + 2 < i && v[2 * j + 2] < v[m]) {
          m = 2 * j + 2;
        }
        if (m == j) {
          return res;
        } else {
          swap(v[j], v[m]);
          j = m;
        }
      }
    }
  };
  int lower_bound(vector<long double>& a, int x) {
    int l = -1, r = a.size();
    int m;
    while (r - l > 1) {
      m = (l + r) / 2;
      if (a[m] < x) {
        l = m;
      } else {
        r = m;
      }
    }
    return r;
  }
  int upper_bound(vector<long double>& a, long double x) {
    int l = -1, r = a.size();
    int m;
    while (r - l > 1) {
      m = (l + r) / 2;
      if (a[m] <= x) {
        l = m;
      } else {
        r = m;
      }
    }
    return l;
  }
  void solve() {
    int n, k, tmp, ans = 0;
    cin >> n >> k;
    vector<pair<int, int> > v;
    for (int i = 0; i < n; ++i) {
      cin >> tmp;
      v.push_back(make_pair(tmp, i));
    }
    sort(v.begin(), v.end());
    for (int i = 0; i < n; ++i) {
      int j = i + 1;
      for (j = i + 1; j < n; ++j) {
        if (v[j].second - v[j - 1].second == 1) {
          continue;
        } else {
          break;
        }
      }
      i = j - 1;
      ans++;
    }
    if (ans <= k) {
      cout << "Yes" << endl;
    } else {
      cout << "No" << endl;
    }
  }
  int main() {
    int t;
    cin >> t;
    for (int i = 0; i < t; ++i) {
      solve();
    }
    return 0;
  }
  