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: import java.io.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  import java.io.IOException;
  import java.io.BufferedReader;
  import java.io.FileReader;
  import java.io.InputStreamReader;
  import java.io.InputStream;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   *
   * @author Omar Khair
   */
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          Scanner in = new Scanner(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          EFixedPoints solver = new EFixedPoints();
          solver.solve(1, in, out);
          out.close();
      }
  
      static class EFixedPoints {
          static int n;
          static int k;
          static int maxDel;
          static int MAXN = 2001;
          static int[] arr;
          static int[][] memo;
  
          public void solve(int testNumber, Scanner sc, PrintWriter out) {
              int tt = sc.nextInt();
              memo = new int[MAXN][MAXN];
              while (tt-- > 0) {
                  n = sc.nextInt();
                  k = sc.nextInt();
                  arr = sc.nextIntArr(n);
                  // bs on number of deletions
                  int ans = -1;
                  int lo = 0, hi = n;
                  while (lo <= hi) {
                      int mid = (lo + hi) / 2;
                      if (poss(mid)) {
                          ans = mid;
                          hi = mid - 1;
                      } else
                          lo = mid + 1;
                  }
                  out.println(ans);
              }
  
          }
  
          private boolean poss(int mid) {
              maxDel = mid;
  
              for (int i = 0; i <= n; i++)
                  for (int j = 0; j <= n; j++)
                      memo[i][j] = -1;
              int maxMatching = dp(0, 0);
  //        System.out.println(mid + " "+ maxMatching);
              return maxMatching >= k;
          }
  
          private int dp(int i, int prevDils) {
              if (i >= n)
                  return 0;
              if (memo[i][prevDils] != -1)
                  return memo[i][prevDils];
              // keep cur element
              int ans = (arr[i] == (i + 1 - prevDils)) ? 1 : 0;
              ans += dp(i + 1, prevDils);
  
              // delete current element
              if (prevDils < maxDel)
                  ans = Math.max(ans, dp(i + 1, prevDils + 1));
              return memo[i][prevDils] = ans;
          }
  
      }
  
      static class Scanner {
          StringTokenizer st;
          BufferedReader br;
  
          public Scanner(FileReader r) {
              br = new BufferedReader(r);
          }
  
          public Scanner(InputStream s) {
              br = new BufferedReader(new InputStreamReader(s));
          }
  
          public String next() {
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return st.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public int[] nextIntArr(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextInt();
              return a;
          }
  
      }
  }
  
  