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;
  void err(const istream_iterator<string>& it) {}
  template <typename T, typename... Args>
  void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << '\n';
    err(++it, args...);
  }
  long double eps = (long double)1 / 1e6;
  long long inf_ll = 1e18 + 1, mod1 = 1e9 + 7, mod2 = 998244353;
  long long mersen_mod = 2305843009213693951;
  long long sqr(long long a) { return a * a; }
  long long qb(long long a) { return a * a * a; }
  long long gcd(long long a, long long b) { return !a ? b : gcd(b % a, a); }
  long long add(long long a, long long b, long long mod) {
    a += b;
    if (a >= mod) a -= mod;
    return a;
  }
  long long sub(long long a, long long b, long long mod) {
    a -= b;
    if (a < 0) a += mod;
    return a;
  }
  long long binpow(long long a, long long b, long long mod) {
    return b ? (b % 2 ? (a * (sqr(binpow(a, b / 2, mod)) % mod)) % mod
                      : sqr(binpow(a, b / 2, mod)) % mod)
             : 1;
  }
  long long binmult(long long a, long long b, long long mod) {
    return b ? (b % 2 ? (2 * binmult(a, b / 2, mod) + a) % mod
                      : (2 * binmult(a, b / 2, mod)) % mod)
             : 0;
  }
  int a[2001], dp[2001][2001], kl[2001][2001], mx[2001][2001];
  int main() {
    cout.tie(0), cin.tie(0), ios_base::sync_with_stdio(0);
    int q, n, i, j, x;
    cin >> q;
    while (q--) {
      cin >> n >> x;
      long long ans = -1;
      for (i = 1; i <= n; i++) cin >> a[i];
      for (i = 1; i <= n; i++)
        for (j = 1; j <= n; j++) dp[i][j] = -1, kl[i][j] = 0;
      for (i = 1; i <= n; i++) {
        for (j = 1; j <= i; j++) {
          if (j == 1) {
            if (a[i] == 1)
              dp[i][j] = 1, kl[i][j] = i - 1;
            else
              dp[i][j] = 0, kl[i][j] = i - 1;
          } else {
            int tmp;
            if (a[i] == j)
              tmp = 1;
            else
              tmp = 0;
            int k = mx[i - 1][j - 1];
            if (dp[k][j - 1] + tmp > dp[i][j]) {
              dp[i][j] = dp[k][j - 1] + tmp;
              kl[i][j] = kl[k][j - 1] + i - k - 1;
            } else if (dp[k][j - 1] + tmp == dp[i][j])
              kl[i][j] = min(kl[i][j], kl[k][j - 1] + i - k - 1);
          }
          if (dp[i][j] >= x && (ans == -1 || ans > kl[i][j])) ans = kl[i][j];
        }
        for (j = 1; j <= i; j++) {
          if (j == 1)
            mx[i][1] = i;
          else {
            mx[i][j] = mx[i - 1][j];
            if (dp[i][j] >= dp[mx[i][j]][j]) mx[i][j] = i;
          }
        }
      }
      cout << ans << '\n';
    }
    return 0;
  }
  