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: // package codeforce.cf734;
  
  import java.io.PrintWriter;
  import java.util.Arrays;
  import java.util.Scanner;
  
  public class E {
      //    MUST SEE BEFORE SUBMISSION
  //    check whether int part would overflow or not, especially when it is a * b!!!!
  
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
          int t = sc.nextInt();
  //        int t = 1;
          for (int i = 0; i < t; i++) {
              solve(sc, pw);
          }
          pw.close();
      }
  
      static void solve(Scanner in, PrintWriter out){
          int n = in.nextInt(), k = in.nextInt();
          int[] arr = new int[n];
          for (int i = 0; i < n; i++) {
              arr[i] = in.nextInt();
          }
  
          int l = 0, r = n + 1;
          while (l < r){
              int[][] dp = new int[n + 1][n + 1];
              for (int i = 0; i < dp.length; i++) {
                  Arrays.fill(dp[i], -1);
              }
              int mid = (l + r) / 2;
              if (dfs(0, 0, arr, dp, mid) >= k){
                  r = mid;
              }else{
                  l = mid + 1;
              }
          }
          if (r == n + 1){
              out.println(-1);
          }else{
              out.println(l);
          }
      }
  
      static int dfs(int idx, int dle, int[] arr, int[][] dp, int tar){
          if (idx == arr.length) return 0;
          if (dp[idx][dle] != -1) return dp[idx][dle];
          int res = 0;
          if (idx + 1 - dle == arr[idx]){
              res = Math.max(res, dfs(idx + 1, dle, arr, dp, tar) + 1);
          }else{
              res = Math.max(res, dfs(idx + 1, dle, arr, dp, tar));
          }
          if (dle < tar) res = Math.max(res, dfs(idx + 1, dle + 1, arr, dp, tar));
          return dp[idx][dle] = res;
      }
  }
  