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.*;
  
  public class CF1 {
      public static void main(String[] args) {
          FastScanner sc=new FastScanner();
          int T=sc.nextInt();
  //        int T=1;
          for (int tt=0; tt<T; tt++){
              int n = sc.nextInt();
              boolean hasIt[]= new boolean [n+1];
              int arr[]= sc.readArray(n);
              sort(arr);
              int i=1;
              int j=0;
              int count=0;
              boolean canBeDone=true;
              while (i<=n){
                  if (hasIt[i]) {i++;continue;}
                  else if (arr[j]<=n && !hasIt[arr[j]]){
                      hasIt[arr[j]]=true;
                      count++;
                      j++;
                  }
                  else {
                      int x=0;
                      if (arr[j]%2==0) x=arr[j]/2-1;
                      else x=arr[j]/2;
                      if (i<=x){
                          hasIt[i]=true;
                          i++;
                          j++;
                      }
                      else {
                          canBeDone =false;
                          break;
                      }
                  }
              }
              if (canBeDone) System.out.println(n-count);
              else System.out.println(-1);
          }
  // arr[i+1]= arr[0]+sum-n*x;
      }
  
      static long mod =998244353L;
      static long power (long a, long b){
          long res=1;
          while (b>0){
              if ((b&1)== 1){
                  res= (res * a % mod)%mod;
              }
              a=(a%mod * a%mod)%mod;
              b=b>>1;
          }
          return res;
      }
      boolean[] sieveOfEratosthenes(int n)
      {
  
          boolean prime[] = new boolean[n+1];
          for(int i=0;i<=n;i++)
              prime[i] = true;
  
          for(int p = 2; p*p <=n; p++)
          {
              if(prime[p] == true)
              {
                  for(int i = p*p; i <= n; i += p)
                      prime[i] = false;
              }
          }
          return prime;
      }
      static void sort(int[] a) {
          ArrayList<Integer> l=new ArrayList<>();
          for (int i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static void sortLong(long[] a) {
          ArrayList<Long> l=new ArrayList<>();
          for (long i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static long gcd (long n, long m){
          if (m==0) return n;
          else return gcd(m, n%m);
      }
  
      static class Pair{
          int x,y;
          public Pair(int x, int y){
              this.x = x;
              this.y = y;
          }
      }
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  }