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;
  long long int pw(long long int a, long long int b, long long int m) {
    long long int ans = 1;
    while (b > 0) {
      if (b & 1) ans = (ans * a) % m;
      a = (a * a) % m;
      b >>= 1;
    }
    return ans % m;
  }
  long long int gcd(long long int a, long long int b) {
    if (a == 0) return b;
    return gcd(b % a, a);
  }
  bool comp(const pair<int, int>& a, const pair<int, int>& b) {
    if (a.second != b.second) {
      return a.second > b.second;
    } else {
      return a.first > b.first;
    }
  }
  double rnd(double var) {
    double value = (long long int)(var * 100 + .5);
    return (double)value / 100;
  }
  long long int const N = (1e6 + 9);
  long long int primes[N + 1];
  void sieve(vector<long long int>& ans) {
    memset(primes, 0, sizeof(primes));
    primes[2] = 1;
    for (long long int i = 3; i <= N; i += 2) primes[i] = 1;
    for (long long int i = 3; i <= N; i += 2) {
      if (primes[i] == 1) {
        for (long long int j = i * i; j <= N; j += i) {
          primes[j] = 0;
        }
      }
    }
    for (long long int i = 2; i <= N; i++) {
      if (primes[i] == 1) {
        ans.push_back(i);
      }
    }
  }
  long long int binarysearch(vector<long long int>& arr, long long int p) {
    long long int l = 0, r = arr.size() - 1;
    long long int mid = 0;
    while (l < r) {
      mid = l + (r - l) / 2;
      if (arr[mid] * arr[mid] == p) {
        return mid;
      } else if (arr[mid] * arr[mid] > p) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return -1;
  }
  void dfs(long long int i, map<long long int, vector<long long int>>& hash,
           vector<long long int>& visited, vector<long long int>& temp) {
    visited[i] = 1;
    temp.push_back(i);
    for (auto x : hash[i]) {
      if (visited[x] == 0) {
        dfs(x, hash, visited, temp);
      }
    }
  }
  int main() {
    long long int t;
    cin >> t;
    while (t--) {
      long long int n, k;
      cin >> n >> k;
      vector<pair<long long int, long long int>> v;
      for (long long int i = 1; i <= n; i++) {
        long long int p;
        cin >> p;
        v.push_back({p, i});
      }
      sort(v.begin(), v.end());
      long long int min_k = 1;
      for (long long int i = 0; i < n - 1; i++) {
        if (v[i + 1].second != 1 + v[i].second) {
          min_k++;
        }
      }
      if (min_k <= k) {
        cout << "YES\n";
      } else {
        cout << "NO\n";
      }
    }
  }
  