You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.

You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:

  * The size of b_i is equal to n for all 1 ≤ i ≤ m. 
  * For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i. 
  * The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m. 



Find the minimum possible value of m, or report that there is no possible m.

Input

The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.

The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).

Output

For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.

Example

Input


6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6


Output


-1
1
2
2
2
1

Note

In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.

In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.

In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m.
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import java.io.*;
import java.util.*;
 
 
public class Main{
   public static void main(String[] args) {
      MyScanner sc = new MyScanner();
      out = new PrintWriter(new BufferedOutputStream(System.out));
      
      // Start writing your solution here. -------------------------------------
      int e=sc.nextInt();
      for (int i=0;i<e;i++) {
          int n=sc.nextInt();
          int k=sc.nextInt();
          int last=-1;
          int x=0;
          for (int j=0;j<n;j++) {
              int temp=sc.nextInt();
              if (temp!=last) {
                  x+=1;
              }
              last=temp;
          }
          if (x==1) {
              out.println("1");
          }
          else if (k==1) {
              out.println("-1");
          }
          else {
              double bleh=(double) (x-1);
              int yeet=(int) (Math.ceil(bleh/(k-1)));
              out.println(yeet);
          }
      }
      /*
      int n      = sc.nextInt();        // read input as integer
      long k     = sc.nextLong();       // read input as long
      double d   = sc.nextDouble();     // read input as double
      String str = sc.next();           // read input as String
      String s   = sc.nextLine();       // read whole line as String

      int result = 3*n;
      out.println(result);                    // print via PrintWriter
      */

      // Stop writing your solution here. -------------------------------------
      out.close();
   }

     

   //-----------PrintWriter for faster output---------------------------------
   public static PrintWriter out;
      
   //-----------MyScanner class for faster input----------
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;
 
      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }
 
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
 
      int nextInt() {
          return Integer.parseInt(next());
      }
 
      long nextLong() {
          return Long.parseLong(next());
      }
 
      double nextDouble() {
          return Double.parseDouble(next());
      }
 
      String nextLine(){
          String str = "";
	  try {
	     str = br.readLine();
	  } catch (IOException e) {
	     e.printStackTrace();
	  }
	  return str;
      }

   }
   //--------------------------------------------------------
}


