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;
  long long mod1 = 998244353, mod2 = 1000000007, limit = 9223372036854775807;
  long double pi = 3.1415926535897932;
  long long modpow(long long x, long long n, long long m) {
    if (x > m) {
      x %= m;
    }
    if (n == 0) return 1 % m;
    long long u = modpow(x, n / 2, m);
    u = (u * u) % m;
    if (n % 2 == 1) u = (u * x) % m;
    return u;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  bool isprime(long long n) {
    if (n == 2) return true;
    for (long long i = 2; i * i <= n; i++) {
      if (n % i == 0) return false;
    }
    return true;
  }
  long long power(long long x, long long n) {
    if (n < 0) {
      return 0;
    }
    long long x_n = 1;
    for (long long i = 0; i < n; i++) {
      x_n *= x;
    }
    return x_n;
  }
  int n, k, mid;
  int arr[2005];
  int dp[2005][2005];
  int getans(int pos, int index) {
    if (pos == n) {
      if (n - index <= mid) {
        return 0;
      }
      return INT_MIN;
    }
    if (dp[pos][index] != -1) {
      return dp[pos][index];
    }
    int ans = INT_MIN;
    if (getans(pos + 1, index) != INT_MIN) {
      ans = getans(pos + 1, index);
    }
    int score = 0;
    if (arr[pos] == index + 1) {
      score = 1;
    }
    if (getans(pos + 1, index + 1) != INT_MIN) {
      ans = max(ans, score + getans(pos + 1, index + 1));
    }
    return dp[pos][index] = ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (int i = 0; i < n; i++) {
        cin >> arr[i];
      }
      int l = 0, r = n - 1;
      int ans = -1;
      while (l <= r) {
        mid = (l + r) / 2;
        for (int i = 0; i <= n; i++) {
          for (int j = 0; j <= n; j++) {
            dp[i][j] = -1;
          }
        }
        if (getans(0, 0) >= k) {
          ans = mid;
          r = mid - 1;
        } else {
          l = mid + 1;
        }
      }
      cout << ans << endl;
    }
  }
  