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;
  using vi = vector<int>;
  using vb = vector<bool>;
  using vs = vector<string>;
  using vc = vector<char>;
  using vd = vector<double>;
  using pii = pair<int, int>;
  using pss = pair<string, string>;
  using ll = long long;
  using tup = tuple<ll, int, int>;
  const int INF = 1e9 + 7;
  const int MOD = 1e9 + 7;
  template <typename Os, class V>
  Os& operator<<(Os& os, V const& v) {
    for (auto const& e : v) os << e << " ";
    return os;
  }
  int N, K, ans = INF;
  const int MAXN = 2001;
  int memo[MAXN], history[MAXN];
  vi nums;
  void solve() {
    cin >> N >> K;
    nums = vi(N + 1, 0);
    for (int n = (1); n <= (N); ++n) cin >> nums[n];
    memset(memo, -1, sizeof memo);
    memset(history, -1, sizeof history);
    ans = INF;
    memo[0] = 0;
    for (int i = (1); i <= (N); ++i) {
      for (int j = (0); j < (i); ++j) {
        if (memo[j] == -1) continue;
        if (nums[j] < nums[i] && (nums[i] - nums[j] - 1 <= i - j - 1) &&
            memo[i] < memo[j] + 1) {
          memo[i] = memo[j] + 1;
          history[i] = j;
        }
      }
    }
    int ans = INF;
    vi deletion(N + 1, 0);
    for (int i = (1); i <= (N); ++i) {
      if (history[i] == -1) continue;
      int j = history[i];
      deletion[i] = deletion[j] + (i - j - 1) - (nums[i] - nums[j] - 1);
      if (memo[i] >= K) ans = min(ans, deletion[i]);
    }
    cout << (ans == INF ? -1 : ans) << "\n";
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    for (int TC = (0); TC < (T); ++TC) {
      solve();
    }
  }
  