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 Mod = 1e9 + 7;
  long long poww(long long a, long long b, long long mod) {
    long long res = 1;
    if (b < 0) b = (b % (mod - 1) + mod - 1) % (mod - 1);
    for (; b; b >>= 1, a = 1ll * a * a % mod)
      if (b & 1) res = 1ll * res * a % mod;
    return res;
  }
  void Max(long long& x, long long y) { x = max(x, y); }
  void Min(long long& x, long long y) { x = min(x, y); }
  void OK(bool yes) { cout << (yes ? "Yes\n" : "No\n"); }
  const long long N = 200500, M = 2010, LN = (1 << 24), K = 17, Mx = 4e6 + 50,
                  inf = 3e18, Mod1 = 1e9 + 7, Mod2 = 999997457;
  const long double pi = acos(-1), eps = 1e-12;
  long long Fact[N];
  void init() {
    Fact[0] = 1ll;
    for (int i = 1; i < N; i++) Fact[i] = (1ll * Fact[i - 1] * i) % Mod;
  }
  int Cnk(long long x, long long y) {
    if (y > x) return 0;
    return (((Fact[x] * poww(Fact[y], -1, Mod)) % Mod) *
            poww(Fact[x - y], -1, Mod)) %
           Mod;
  }
  void Add(long long& x, long long y, int j = 0) {
    long long mod = Mod;
    if (j == 1) mod = Mod1;
    if (j == 2) mod = Mod2;
    x %= mod;
    y %= mod;
    if (x < 0) x += mod;
    if (y < 0) y += mod;
    x = (x + y >= mod ? x + y - mod : x + y);
  }
  long long dp[2020], a[2020];
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    int T = 1;
    int Case = 1;
    init();
    cin >> T;
    while (T--) {
      int n, k;
      cin >> n >> k;
      for (int i = 0; i <= n; i++) dp[i] = 0;
      for (int i = 1; i <= n; i++) {
        cin >> a[i];
        if (i - a[i] >= 0) dp[i] = 1;
      }
      long long Ans = inf;
      if (k == 1) {
        for (int i = 1; i <= n; i++)
          if (a[i] <= i) Min(Ans, i - a[i]);
        if (Ans == inf)
          cout << -1 << '\n';
        else
          cout << Ans << '\n';
        continue;
      }
      for (int i = 1; i <= n; i++) {
        if (i - a[i] < 0) continue;
        for (int j = 1; j < i; j++) {
          if (j - a[j] < 0) continue;
          if (i - a[i] >= j - a[j] && a[i] > a[j]) {
            Max(dp[i], dp[j] + 1);
            if (dp[i] >= k) {
              Min(Ans, i - a[i]);
            }
          }
        }
      }
      if (Ans < inf)
        cout << Ans << '\n';
      else
        cout << -1 << '\n';
    }
    return 0;
  }
  