description: Paprika loves permutations. She has an array a_1, a_2, ..., a_n. She wants to make the array a permutation of integers 1 to n.
  
  In order to achieve this goal, she can perform operations on the array. In each operation she can choose two integers i (1 ≤ i ≤ n) and x (x > 0), then perform a_i := a_i mod x (that is, replace a_i by the remainder of a_i divided by x). In different operations, the chosen i and x can be different.
  
  Determine the minimum number of operations needed to make the array a permutation of integers 1 to n. If it is impossible, output -1.
  
  A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
  
  Input
  
  Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
  
  The first line of each test case contains an integer n (1 ≤ n ≤ 10^5).
  
  The second line of each test case contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^9).
  
  It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case, output the minimum number of operations needed to make the array a permutation of integers 1 to n, or -1 if it is impossible.
  
  Example
  
  Input
  
  
  4
  2
  1 7
  3
  1 5 4
  4
  12345678 87654321 20211218 23571113
  9
  1 2 3 4 18 19 5 6 7
  
  
  Output
  
  
  1
  -1
  4
  2
  
  Note
  
  For the first test, the only possible sequence of operations which minimizes the number of operations is: 
  
    * Choose i=2, x=5. Perform a_2 := a_2 mod 5 = 2. 
  
  
  
  For the second test, it is impossible to obtain a permutation of integers from 1 to n.
solution: import java.util.ArrayList;
  import java.util.Collections;
  import java.util.HashSet;
  import java.util.Scanner;
  
  /**
   *
   * @author Acer
   */
  public class PaprikaAndPermutation1_C {
      
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int T = sc.nextInt();
          while(T-- > 0){
              int n = sc.nextInt();
              ArrayList<Integer> arr = new ArrayList<>();
              for (int i = 0; i < n; i++) {
                  arr.add(sc.nextInt());
              }
              //Collections.sort(A);
              ArrayList<Integer> A = new ArrayList<>();            
              HashSet<Integer> set = new HashSet<>();
              for (int i = 0; i < n; i++) {
                  if(set.contains(arr.get(i)) || arr.get(i) > n){
                      A.add(arr.get(i));
                  }
                  else{
                      set.add(arr.get(i));
                  }
              }
              ArrayList<Integer> B = new ArrayList<>();
              for (int i = 1; i <= n; i++) {
                  if(!set.contains(i)){
                      B.add(i);
                  }
              }
              
              if(A.isEmpty()){
                  System.out.println(0);
              }
              else{
                  Collections.sort(A);
                  int j = 0;
                  boolean flag = true;
                  for (int i = 0; i < A.size(); i++) {
                      if((A.get(i)+1)/2 <= B.get(j)){
                          flag = false;
                          break;
                      }
                      j++;
                  }
                  
                  if(flag){
                      System.out.println(A.size());
                  }
                  else{
                      System.out.println(-1);
                  }
              }
          }
      }
      
  }