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.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.Comparator;
  import java.util.PriorityQueue;
  import java.util.StringTokenizer;
  
  public class C {
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader()
          {
              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;
          }
      }
      public static void main(String[] args){
          FastReader sc = new FastReader();
          int t = sc.nextInt();
          StringBuilder s = new StringBuilder();
          while (t-->0){
              int n = sc.nextInt();
              int[] a = new int[n];
              boolean[] p = new boolean[n+1];
              for (int i=1;i<=n;i++){
                  p[i] = false;
              }
              PriorityQueue<Integer> maxHeap = new PriorityQueue<>(new Comparator<Integer>() {
                  @Override
                  public int compare(Integer o1, Integer o2) {
                      return o2-o1;
                  }
              });
              for (int i=0;i<n;i++){
                  a[i] = sc.nextInt();
                  if (a[i]<=n){
                      if (p[a[i]]){
                          maxHeap.add(a[i]%((a[i]/2)+1));
                      }else {
                          p[a[i]] = true;
                      }
                  }else {
                      maxHeap.add(a[i]%((a[i]/2)+1));
                  }
              }
  
              int count = 0;
              for (int i=n;i>0;i--){
                  if (!p[i]){
                      if (maxHeap.peek()<i){
                          count = -1;
                          break;
                      }else {
                          maxHeap.poll();
                          count++;
                      }
                  }
              }
              s.append(count+"\n");
  
          }
          System.out.println(s);
      }
  }
  