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 maxs = 1e6 + 5;
  long long root[maxs];
  class cmp {
    bool operator()(const long long &a, const long long &b) { return a < b; }
  };
  void init() { iota(root, root + maxs, 0); }
  long long find(long long u) {
    long long f;
    if (root[u] == u)
      return u;
    else {
      long long f = find(root[u]);
      root[u] = f;
      return f;
    }
  }
  void Union(long long x, long long y) {
    long long u = find(x);
    long long v = find(y);
    root[v] = u;
  }
  long long power(long long x, long long y, long long p) {
    long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  void update(vector<long long> &BIT, long long ind, long long val) {
    while (ind < BIT.size()) {
      BIT[ind] += val;
      ind += (ind & (-ind));
    }
  }
  long long query(vector<long long> &BIT, long long ind) {
    long long sum = 0;
    while (ind > 0) {
      sum += BIT[ind];
      ind -= (ind & (-ind));
    }
    return sum;
  }
  void count_factors() {
    long long i, j;
    long long numOfDivisors[maxs] = {0};
    long long ans[maxs] = {0};
    for (i = 1; i < maxs; i++)
      for (j = i; j < maxs; j += i) {
        numOfDivisors[j]++;
        ans[j]++;
      }
    for (i = 1; i < maxs; i++)
      for (j = i; j < maxs; j += i)
        if (numOfDivisors[j / i] == 4) ans[j] = min(ans[j], ans[i]);
  }
  long long get(long long x, long long y) {
    if (x == 0) return 1;
    long long cnt = 0;
    while (x % y == 0) {
      cnt++;
      x /= y;
    }
    return cnt;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t, n, i, k, y, x, j, m, r, l;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      vector<long long> arr(n + 1);
      for (i = 1; i <= n; i++) {
        cin >> arr[i];
      }
      vector<vector<long long>> dp(n + 1, vector<long long>(n + 1, 0));
      for (i = 1; i <= n; i++) {
        for (j = 1; j <= i; j++) {
          dp[i][j] = dp[i - 1][j];
          dp[i][j] = max(dp[i - 1][j - 1] + (arr[i] == j), dp[i][j]);
        }
      }
      long long ans = 0;
      for (i = 1; i <= n; i++) {
        if (dp[n][i] >= k) ans = i;
      }
      if (ans == 0)
        cout << -1 << '\n';
      else
        cout << n - ans << '\n';
    }
    return 0;
  }
  