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>
  const long long int mod = 1000000000 + 7;
  const long long int INF = 9000000000000000000;
  using namespace std;
  void yes() { cout << "YES" << endl; }
  void no() { cout << "NO" << endl; }
  long long int gcd(long long int a, long long int b) {
    return (a == 0 ? b : gcd(b % a, a));
  }
  long long int lcm(long long int a, long long int b) {
    return (a * b / (gcd(a, b)));
  }
  long long int add(long long int a, long long int b, long long int c = 0) {
    return ((a % mod + b % mod) % mod + c % mod) % mod;
  }
  long long int mul(long long int a, long long int b) {
    return ((a % mod) * (b % mod)) % mod;
  }
  long long int sub(long long int a, long long int b) {
    return (a % mod - b % mod + mod) % mod;
  }
  void print(vector<long long int>& ans) {
    for (auto x : ans) cout << x << " ";
    cout << endl;
  }
  void print(long long int a) { cout << a << endl; }
  long long int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
  long long int n, w, h, d, k, m, mx, timer, realans, cc, start, queries,
      startindex;
  long long int arr[2005], dp[2005][2005];
  long long int go(long long int id, long long int del) {
    if (id < 0) return 0;
    if (del == 0) {
      long long int cnt = 0;
      for (long long int i = id; i >= 1; i--) {
        if (arr[i] == i) cnt++;
      }
      return cnt;
    }
    if (dp[id][del] != -1) return dp[id][del];
    long long int fst = 0, scd = 0;
    if (arr[id] == id - del) {
      fst = 1 + go(id - 1, del);
    } else {
      fst = go(id - 1, del);
    }
    scd = go(id - 1, del - 1);
    return dp[id][del] = max(fst, scd);
  }
  void solve() {
    cin >> n >> k;
    for (long long int i = 1; i <= n; i++) cin >> arr[i];
    long long int ans = -1;
    memset(dp, -1, sizeof(dp));
    for (long long int i = 0; i <= n; i++) {
      if (go(n, i) >= k) {
        ans = i;
        break;
      }
    }
    cout << ans << endl;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    long long int tests = 1;
    cin >> tests;
    while (tests--) {
      solve();
    }
    return 0;
  }
  