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: /*
   * To change this license header, choose License Headers in Project Properties.
   * To change this template file, choose Tools | Templates
   * and open the template in the editor.
   */
  
  import java.io.BufferedReader;
  import java.io.DataInputStream;
  import java.io.FileInputStream;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  /**
   *
   * @author is2ac
   */
  public class D_CF {
  
      public static void main(String[] args) throws IOException {
          FastScanner58 fs = new FastScanner58();
          //Reader fs = new Reader();
          PrintWriter pw = new PrintWriter(System.out);
          int t = fs.ni();
          //int t = 1;
          for (int tc = 0; tc < t; tc++) {
              int n = fs.ni();
              int k = fs.ni();
              int[] a = fs.intArray(n);
              Integer[][] dp = new Integer[n][k+1];
              int l = 0, r = n;
              while (l<r) {
                  int m = (l+r)/2;
                  dp = new Integer[n][m+1];
                  if (recur(0,0,m,a,dp)>=k) {
                      r = m;
                  } else {
                      l = m + 1;
                  }
              }
              if (l==n) {
                  dp = new Integer[n][n+1];
                  if (recur(0,0,n,a,dp)<k) {
                      pw.println(-1);
                  } else {
                      pw.println(l);
                  }
              } else {
                  pw.println(l);
              }
          }
          pw.close();
      }
      public static int recur(int ind, int rem, int k, int[] a, Integer[][] dp) {
          int n = 0;
          if (rem>k) return (int)(-1e9);
          if (ind==a.length) return 0;
          if (dp[ind][rem]!=null) return dp[ind][rem];
          n = Math.max(n,recur(ind+1,rem+1,k,a,dp));
          int x  = 0;
          if (ind+1-rem==a[ind]) {
              x = 1;
          }
          n = Math.max(n,recur(ind+1,rem,k,a,dp)+x);
          return dp[ind][rem] = n;
      }
      public static boolean isPrime(int n) {
          if (n == 1) {
              return false;
          }
          for (int i = 2; i <= (int) Math.sqrt(n); i++) {
              if (n % i == 0) {
                  return false;
              }
          }
          return true;
      }
  
      public static int primes(int a) {
          int b = a;
          int ct = 0;
          for (int i = 2; i <= (int) Math.sqrt(b); i++) {
              if (a % i != 0) {
                  continue;
              }
              if (!isPrime(i)) {
                  continue;
              }
              int j = a / i;
              while (a % i == 0) {
                  a /= i;
                  ct++;
              }
              if (!isPrime(j)) {
                  continue;
              }
              if (j == 1) {
                  continue;
              }
              while (a % j == 0) {
                  a /= j;
                  ct++;
              }
          }
          return ct;
      }
  
      public static int dis(int a, int b, int n) {
          //for a to get to b
          if (a < b) {
              return b - a;
          }
          b += n;
          return b - a;
      }
  
      public static long gcd(long n1, long n2) {
          if (n2 == 0) {
              return n1;
          }
          return gcd(n2, n1 % n2);
      }
  }
  
  class GFG {
  
      // A utility function to get the
      // middle index of given range.
      static int getMid(int s, int e) {
          return s + (e - s) / 2;
      }
  
      /*
      * A recursive function to get the sum
      of values in given range of the array.
      * The following are parameters
        for this function.
      *
      * st -> Pointer to segment tree
      * node -> Index of current node in
      *         the segment tree.
      * ss & se -> Starting and ending indexes
      *         of the segment represented
      *         by current node, i.e., st[node]
      * l & r -> Starting and ending indexes
      *         of range query
       */
      static long MinUtil(long[] st, int ss,
              int se, int l,
              int r, int node) {
  
          // If segment of this node is completely
          // part of given range, then return
          // the max of segment
          if (l <= ss && r >= se) {
              return st[node];
          }
  
          // If segment of this node does not
          // belong to given range
          if (se < l || ss > r) {
              return (long) (1e11);
          }
  
          // If segment of this node is partially
          // the part of given range
          int mid = getMid(ss, se);
  
          return Math.min(
                  MinUtil(st, ss, mid, l, r,
                          2 * node + 1),
                  MinUtil(st, mid + 1, se, l, r,
                          2 * node + 2));
      }
  
      /*
      * A recursive function to update the
      nodes which have the given index in their
      * range. The following are parameters
      st, ss and se are same as defined above
      * index -> index of the element to be updated.
       */
      static void updateValue(long arr[], long[] st, int ss,
              int se, int index,
              long value,
              int node) {
          if (index < ss || index > se) {
              //System.out.println("Invalid Input");
              return;
          }
  
          if (ss == se) {
  
              // update value in array and in
              // segment tree
              arr[index] = value;
              st[node] = value;
          } else {
              int mid = getMid(ss, se);
  
              if (index >= ss && index <= mid) {
                  updateValue(arr, st, ss, mid,
                          index, value,
                          2 * node + 1);
              } else {
                  updateValue(arr, st, mid + 1, se, index,
                          value, 2 * node + 2);
              }
  
              st[node] = Math.min(st[2 * node + 1],
                      st[2 * node + 2]);
          }
          return;
      }
  
      // Return max of elements in range from
      // index l (query start) to r (query end).
      static long getMin(long[] st, int n, int l, int r) {
  
          // Check for erroneous input values
          if (l < 0 || r > n - 1 || l > r) {
              //System.out.printf("Invalid Input\n");
              return (long) (1e11);
          }
  
          return MinUtil(st, 0, n - 1, l, r, 0);
      }
  
      // A recursive function that constructs Segment
      // Tree for array[ss..se]. si is index of
      // current node in segment tree st
      static long constructSTUtil(long arr[],
              int ss, int se,
              long[] st, int si) {
  
          // If there is one element in array, store
          // it in current node of segment tree and return
          if (ss == se) {
              st[si] = arr[ss];
              return arr[ss];
          }
  
          // If there are more than one elements, then
          // recur for left and right subtrees and
          // store the max of values in this node
          int mid = getMid(ss, se);
  
          st[si] = Math.min(
                  constructSTUtil(arr, ss, mid,
                          st, si * 2 + 1),
                  constructSTUtil(arr, mid + 1,
                          se, st,
                          si * 2 + 2));
  
          return st[si];
      }
  
      /*
      * Function to construct segment tree from
      given array. This function allocates
      * memory for segment tree.
       */
      static long[] constructST(long arr[], int n) {
  
          // Height of segment tree
          int x = (int) Math.ceil(Math.log(n) / Math.log(2));
  
          // Maximum size of segment tree
          int max_size = 2 * (int) Math.pow(2, x) - 1;
  
          // Allocate memory
          long[] st = new long[max_size];
  
          // Fill the allocated memory st
          constructSTUtil(arr, 0, n - 1, st, 0);
  
          // Return the constructed segment tree
          return st;
      }
  }
  
  class BIT16 {
  
      long[] bit;
  
      public BIT16(int size) {
          bit = new long[size];
      }
  
      public void update(int ind, long delta) {
          while (ind < bit.length) {
              bit[ind] += delta;
              ind = ind + (ind & (-1 * ind));
          }
      }
  
      public long sum(int ind) {
          long s = 0;
          while (ind > 0) {
              s += bit[ind];
              ind = ind - (ind & (-1 * ind));
          }
          return s;
      }
  
      public long query(int l, int r) {
          return sum(r) - sum(l);
      }
  }
  
  class UnionFind18 {
  
      int[] id;
  
      public UnionFind18(int size) {
          id = new int[size];
          for (int i = 0; i < size; i++) {
              id[i] = i;
          }
      }
  
      public int find(int p) {
          int root = p;
          while (root != id[root]) {
              root = id[root];
          }
          while (p != root) {
              int next = id[p];
              id[p] = root;
              p = next;
          }
          return root;
      }
  
      public void union(int p, int q) {
          int a = find(p), b = find(q);
          if (a == b) {
              return;
          }
          id[b] = a;
      }
  }
  
  class Reader {
  
      final private int BUFFER_SIZE = 1 << 16;
      private DataInputStream din;
      private byte[] buffer;
      private int bufferPointer, bytesRead;
  
      public Reader() {
          din = new DataInputStream(System.in);
          buffer = new byte[BUFFER_SIZE];
          bufferPointer = bytesRead = 0;
      }
  
      public Reader(String file_name) throws IOException {
          din = new DataInputStream(new FileInputStream(file_name));
          buffer = new byte[BUFFER_SIZE];
          bufferPointer = bytesRead = 0;
      }
  
      public String readLine() throws IOException {
          byte[] buf = new byte[64];
          int cnt = 0, c;
          while ((c = read()) != -1) {
              if (c == '\n') {
                  break;
              }
              buf[cnt++] = (byte) c;
          }
          return new String(buf, 0, cnt);
      }
  
      public int ni() throws IOException {
          int ret = 0;
          byte c = read();
          while (c <= ' ') {
              c = read();
          }
          boolean neg = (c == '-');
          if (neg) {
              c = read();
          }
          do {
              ret = ret * 10 + c - '0';
          } while ((c = read()) >= '0' && c <= '9');
          if (neg) {
              return -ret;
          }
          return ret;
      }
  
      public long nextLong() throws IOException {
          long ret = 0;
          byte c = read();
          while (c <= ' ') {
              c = read();
          }
          boolean neg = (c == '-');
          if (neg) {
              c = read();
          }
          do {
              ret = ret * 10 + c - '0';
          } while ((c = read()) >= '0' && c <= '9');
          if (neg) {
              return -ret;
          }
          return ret;
      }
  
      public double nextDouble() throws IOException {
          double ret = 0, div = 1;
          byte c = read();
          while (c <= ' ') {
              c = read();
          }
          boolean neg = (c == '-');
          if (neg) {
              c = read();
          }
          do {
              ret = ret * 10 + c - '0';
          } while ((c = read()) >= '0' && c <= '9');
          if (c == '.') {
              while ((c = read()) >= '0' && c <= '9') {
                  ret += (c - '0') / (div *= 10);
              }
          }
          if (neg) {
              return -ret;
          }
          return ret;
      }
  
      private void fillBuffer() throws IOException {
          bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
          if (bytesRead == -1) {
              buffer[0] = -1;
          }
      }
  
      private byte read() throws IOException {
          if (bufferPointer == bytesRead) {
              fillBuffer();
          }
          return buffer[bufferPointer++];
      }
  
      public void close() throws IOException {
          if (din == null) {
              return;
          }
          din.close();
      }
  }
  
  class FastScanner58 {
  
      BufferedReader br;
      StringTokenizer st;
  
      public FastScanner58() {
          br = new BufferedReader(new InputStreamReader(System.in), 32768);
          st = null;
      }
  
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
  
      int ni() {
          return Integer.parseInt(next());
      }
  
      int[] intArray(int N) {
          int[] ret = new int[N];
          for (int i = 0; i < N; i++) {
              ret[i] = ni();
          }
          return ret;
      }
  
      long nl() {
          return Long.parseLong(next());
      }
  
      long[] longArray(int N) {
          long[] ret = new long[N];
          for (int i = 0; i < N; i++) {
              ret[i] = nl();
          }
          return ret;
      }
  
      double nd() {
          return Double.parseDouble(next());
      }
  
      String nextLine() {
          String str = "";
          try {
              str = br.readLine();
          } catch (IOException e) {
              e.printStackTrace();
          }
          return str;
      }
  }
  