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;
  const long long MAX_N = 1e5 + 1;
  const long long MOD = 1e9 + 7;
  const long long INF = 1e9;
  const long long LINF = 1e18;
  void solve() {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n);
    for (long long i{}; i < n; ++i) {
      cin >> a[i];
    }
    vector<long long> costs(n);
    for (long long i{}; i < n; ++i) {
      if (a[i] > i + 1)
        costs[i] = INF;
      else
        costs[i] = i + 1 - a[i];
    }
    vector<array<long long, 2>> dp(n);
    for (long long i{}; i < n; ++i) {
      long long largest = 1;
      long long cost = costs[i];
      long long t = -1;
      for (long long j{}; j < i; ++j) {
        if (a[j] >= a[i]) continue;
        if (dp[j][0] > largest - 1) {
          if (costs[i] != INF && costs[i] >= costs[j]) {
            largest = dp[j][0] + 1;
            cost = dp[j][1] + (costs[i] - costs[j]);
            t = j;
          }
        } else if (dp[j][0] == largest - 1) {
          if (costs[i] != INF && costs[i] >= costs[j] &&
              cost > dp[j][1] + (costs[i] - costs[j])) {
            cost = dp[j][1] + (costs[i] - costs[j]);
            t = j;
          }
        }
      }
      dp[i] = {largest, cost};
    }
    long long out = -1;
    for (long long i{}; i < n; ++i) {
      if (dp[i][0] >= k) {
        if (out == -1) {
          out = dp[i][1];
        } else {
          out = min(out, dp[i][1]);
        }
      }
    }
    cout << out << '\n';
  }
  long long main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long tc;
    cin >> tc;
    for (long long t = 1; t <= tc; t++) {
      solve();
    }
  }
  