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;
  void debugs() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  int multiply(long long x, long long res[], long long ressize) {
    long long carry = 0;
    for (int i = 0; i < ressize; i++) {
      int product = res[i] * x + carry;
      res[i] = product % 10;
      carry = product / 10;
    }
    while (carry) {
      res[ressize] = carry % 10;
      carry = carry / 10;
      ressize++;
    }
    return ressize;
  }
  void bigfactorial(long long n) {
    long long ressize = 1;
    long long res[10000];
    res[0] = 1;
    for (int i = 2; i <= n; i++) {
      ressize = multiply(i, res, ressize);
    }
    for (int i = ressize - 1; i >= 0; i--) {
      cout << res[i];
    }
  }
  long long n;
  bool palindrome(string s) {
    string t;
    t = s;
    reverse(t.begin(), t.end());
    if (s == t) {
      return true;
    }
    return false;
  }
  bool issafe(vector<vector<int>> board, int row, int col) {
    for (int i = 0; i < col; i++) {
      if (board[row][i]) return false;
    }
    for (int i = row, j = col; i >= 0 && j >= 0; i--, j--) {
      if (board[i][j] == 1) return false;
    }
    for (int i = row, j = col; i < n && j < n; i++, j++) {
      if (board[i][j] == 1) return false;
    }
    return true;
  }
  void display(vector<vector<int>> &board) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        cout << board[i][j] << " ";
      }
      cout << endl;
    }
  }
  bool solvenqueen(vector<vector<int>> &board, int col) {
    if (col >= n) {
      return true;
    }
    for (int i = 0; i < n; i++) {
      if (issafe(board, i, col)) {
        board[i][col] = 1;
        if (solvenqueen(board, col + 1)) return true;
        board[i][col] = 0;
      }
    }
    return false;
  }
  int main() {
    int n;
    int k;
    cin >> n >> k;
    long long a[n];
    vector<pair<int, int>> v(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      v[i].first = a[i];
      v[i].second = i;
    }
    sort(v.begin(), v.end());
    int ans = n;
    for (int i = 0; i < n - 1; i++) {
      if (v[i + 1].second == v[i].second + 1) {
        ans -= 1;
      }
    }
    if (ans <= k) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  }
  