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>
  #pragma GCC optimize("O3")
  using namespace std;
  template <typename... Args>
  void logger(string vars, Args &&...values) {
    cout << vars << " = ";
    string delim = "";
  }
  inline string toString(int x) {
    string ans = "";
    bool negtive = x < 0;
    x = abs(x);
    while (x) ans.push_back(x % 10 + '0'), x /= 10;
    if (ans.size() == 0) ans = "0";
    if (negtive) ans.push_back('-');
    for (int i = 0; i < ans.size() / 2; i++)
      swap(ans[i], ans[ans.size() - 1 - i]);
    return ans;
  }
  inline int toInt(const string &str) {
    int ans = 0;
    bool negtive = str[0] == '-';
    for (int i = negtive; i < str.size(); i++) ans = ans * 10 + str[i] - '0';
    if (negtive) ans *= -1;
    return ans;
  }
  inline int gcd(int x, int y) { return x ? gcd(y % x, x) : y; }
  long long exgcd(long long a, long long b, long long &x, long long &y) {
    long long d = a;
    if (b == 0)
      x = 1, y = 0;
    else {
      d = exgcd(b, a % b, y, x), y -= a / b * x;
    }
    return d;
  }
  const int mod = 998244353;
  inline int read() {
    int ans = 0;
    char last = ' ', ch = getchar();
    while (!(ch >= '0' && ch <= '9')) last = ch, ch = getchar();
    while (ch >= '0' && ch <= '9')
      ans = (ans << 3) + (ans << 1) + ch - '0', ch = getchar();
    return ans;
  }
  inline int QPow(long long a, long long b) {
    long long ans = 1;
    if (b < 0) return 0;
    while (b) {
      if (b & 1) ans = ans * a;
      a = a * a;
      b >>= 1;
    }
    return ans;
  }
  inline int QPow(long long a, long long b, const int &mod) {
    long long ans = 1;
    while (b) {
      if (b & 1) ans = ans * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  const double PI = 3.141592653589793115997963468544185161590576171875;
  const double e = 2.718281828459045;
  void ask(int x) {
    cout << '?' << ' ' << x << endl;
    cout.flush();
    ;
  }
  void decide(int x) {
    cout << "! " << x << endl;
    cout.flush();
    ;
  }
  short _min(short a, short b) { return a < b ? a : b; }
  const int inf = (long long)1 << 60;
  const int inf2 = 10000;
  const double eps = 1e-10;
  int db_cmp(double x, double y) {
    if (fabs(x - y) < eps) return 0;
    if (x > y) return 1;
    return -1;
  }
  const int maxn = 2002;
  int arr[maxn];
  int dp[maxn][maxn];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t, n, k;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (int i = 1; i <= n; i++) cin >> arr[i];
      dp[1][1] = 0;
      dp[1][0] = arr[1] == 1;
      for (int i = 2; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
          dp[i][j] = max(dp[i - 1][j - 1], dp[i - 1][j] + (arr[i] == i - j));
        }
        dp[i][0] = (arr[i] == i) + dp[i - 1][0];
        dp[i][i] = 0;
      }
      int ans = -1;
      for (int i = 0; i <= n; i++)
        if (dp[n][i] >= k) {
          ans = i;
          break;
        }
      cout << ans << '\n';
      for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) dp[i][j] = 0;
    }
    return 0;
  }
  