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 int N = 2e3 + 10;
  int a[N];
  int b[N];
  int dp[N][N];
  void work() {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) b[i] = 0;
    for (int i = 1; i <= n; i++) cin >> a[i];
    dp[0][0] = 0;
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= n; j++) dp[i][j] = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] == i)
        dp[i][0] = dp[i - 1][0] + 1;
      else
        dp[i][0] = dp[i - 1][0];
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j < i; j++) {
        dp[i][j] = max(dp[i - 1][j] + (a[i] + j == i), dp[i - 1][j - 1]);
      }
      dp[i][i] = 0;
    }
    int ans = n + 1;
    for (int j = 0; j <= n; j++) {
      if (dp[n][j] >= k) ans = min(ans, j);
    }
    if (ans == n + 1)
      cout << "-1" << endl;
    else
      cout << ans << endl;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      work();
    }
  }
  