description: Consider a sequence of integers a_1, a_2, …, a_n. In one move, you can select any element of the sequence and delete it. After an element is deleted, all elements to the right are shifted to the left by 1 position, so there are no empty spaces in the sequence. So after you make a move, the sequence's length decreases by 1. The indices of the elements after the move are recalculated.
  
  E. g. let the sequence be a=[3, 2, 2, 1, 5]. Let's select the element a_3=2 in a move. Then after the move the sequence will be equal to a=[3, 2, 1, 5], so the 3-rd element of the new sequence will be a_3=1 and the 4-th element will be a_4=5.
  
  You are given a sequence a_1, a_2, …, a_n and a number k. You need to find the minimum number of moves you have to make so that in the resulting sequence there will be at least k elements that are equal to their indices, i. e. the resulting sequence b_1, b_2, …, b_m will contain at least k indices i such that b_i = i.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
  
  Each test case consists of two consecutive lines. The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2000). The second line contains a sequence of integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). The numbers in the sequence are not necessarily different.
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 2000.
  
  Output
  
  For each test case output in a single line:
  
    * -1 if there's no desired move sequence; 
    * otherwise, the integer x (0 ≤ x ≤ n) — the minimum number of the moves to be made so that the resulting sequence will contain at least k elements that are equal to their indices. 
  
  Example
  
  Input
  
  
  4
  7 6
  1 1 2 3 4 5 6
  5 2
  5 1 3 2 3
  5 2
  5 5 5 5 4
  8 4
  1 2 3 3 2 2 5 5
  
  
  Output
  
  
  1
  2
  -1
  2
  
  Note
  
  In the first test case the sequence doesn't satisfy the desired condition, but it can be provided by deleting the first element, hence the sequence will be [1, 2, 3, 4, 5, 6] and 6 elements will be equal to their indices.
  
  In the second test case there are two ways to get the desired result in 2 moves: the first one is to delete the 1-st and the 3-rd elements so that the sequence will be [1, 2, 3] and have 3 elements equal to their indices; the second way is to delete the 2-nd and the 3-rd elements to get the sequence [5, 2, 3] with 2 desired elements.
solution: #include <bits/stdc++.h>
  using namespace std;
  template <typename T>
  istream &operator>>(istream &is, vector<T> &v) {
    for (auto &it : v) is >> it;
    return is;
  }
  template <typename T>
  ostream &operator<<(ostream &os, const vector<T> &v) {
    os << '{';
    string sep = "";
    for (const auto &x : v) os << sep << x, sep = ", ";
    return os << '}';
  }
  template <typename A, typename B>
  ostream &operator<<(ostream &os, const pair<A, B> &p) {
    return os << '(' << p.first << ", " << p.second << ')';
  }
  void dbg_out() { cerr << " end.\n"; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cerr << H << ' ';
    dbg_out(T...);
  }
  const int N = 2e3 + 5;
  const int INF = 0x3f3f3f3f;
  const int mod = 1e9 + 7;
  int dp[N];
  int arr[N];
  void solve(int T) {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) {
      dp[i] = -1;
      cin >> arr[i];
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = i; j >= 1; --j) {
        dp[j] = max(dp[j], dp[j - 1] + (arr[i] == j));
      }
    }
    for (int i = n; i >= 1; --i) {
      if (dp[i] >= k) {
        cout << n - i << "\n";
        return;
      }
    }
    cout << "-1\n";
  }
  int main() {
    cin.tie(0), ios::sync_with_stdio(0);
    int t;
    cin >> t;
    for (int tt = 0; tt < t; ++tt) solve(1);
    return 0;
  }
  