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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
  #pragma GCC optimize("-ffloat-store")
  using namespace std;
  const auto start_time = std::chrono::high_resolution_clock::now();
  void aryanc403() {}
  const long long int INF = 0xFFFFFFFFFFFFFFFL;
  long long int seed;
  mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
  inline long long int rnd(long long int l = 0, long long int r = INF) {
    return uniform_int_distribution<long long int>(l, r)(rng);
  }
  class CMP {
   public:
    bool operator()(pair<long long int, long long int> a,
                    pair<long long int, long long int> b) {
      return !(a.first < b.first || (a.first == b.first && a.second <= b.second));
    }
  };
  void add(map<long long int, long long int> &m, long long int x,
           long long int cnt = 1) {
    auto jt = m.find(x);
    if (jt == m.end())
      m.insert({x, cnt});
    else
      jt->second += cnt;
  }
  void del(map<long long int, long long int> &m, long long int x,
           long long int cnt = 1) {
    auto jt = m.find(x);
    if (jt->second <= cnt)
      m.erase(jt);
    else
      jt->second -= cnt;
  }
  bool cmp(const pair<long long int, long long int> &a,
           const pair<long long int, long long int> &b) {
    return a.first < b.first || (a.first == b.first && a.second < b.second);
  }
  const long long int mod = 1000000007L;
  long long int T, n, i, j, k, in, cnt, l, r, u, v, x, y;
  long long int m;
  string s;
  vector<long long int> a;
  vector<vector<long long int> > dp;
  long long int solve(long long int n, long long int k) {
    if (k == -1) return 0;
    if (n == -1) {
      return -INF;
    }
    42;
    ;
    long long int &ans = dp[n][k];
    if (ans != -1) return ans;
    ans = -INF;
    ans = max(ans, solve(n - 1, k));
    ans = max(ans, solve(n - 1, k - 1) + (a[n] == k));
    return ans;
  }
  int main(void) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> T;
    while (T--) {
      cin >> n >> k;
      a.clear();
      a.reserve(n);
      for (i = 0; i < (n); ++i) {
        cin >> in;
        in--;
        a.push_back(in);
      }
      dp.clear();
      dp.resize(n, vector<long long int>(n, -1));
      long long int ans = -1;
      for (long long int i = n - 1; i >= 0; --i) {
        const long long int c = solve(n - 1, i);
        if (c >= k) {
          ans = n - i - 1;
          break;
        }
      }
      cout << ans << "\n";
    }
    aryanc403();
    return 0;
  }
  