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.*;
  public class Codeforces {
  
  
  	// CHECK CONSTRAINTS ALWAYS EDGE CASE MISSED 
  	
  static int mod = 1000000007;	
  
  static class Pair{
  	long a;
  	long b;
  	Pair(long a,  long b){
  		this.a = a;
  		this.b = b;
  	}
  }
  
  public static void main(String[] args) {
  
  Scanner sc=  new Scanner(System.in);
  
    while(sc.hasNext()) {
  
  int t  =  sc.nextInt();
   while(t-->0) {
  	
  int n  = sc.nextInt();
  ArrayList<Integer> al = new ArrayList<>();
  ArrayList<Integer> bl = new ArrayList<>();
   int arr[] = new int[n+1];
  for(int i = 0 ; i <n;i++) {
  	int a = sc.nextInt();
  	if(a>n) {
  		al.add(a);
  	}
  	else {
  		arr[a]++;
  	}
  }
  
  for(int i =1 ; i<=n ; i++) {
  	if(arr[i] == 0) {
  		bl.add(i);
  	}
  	if(arr[i]>1) {
  		while(arr[i] -- >1) {
  			al.add(i);
  			}
  		
  	}
  }
  Collections.sort(al);
  Collections.sort(bl);
  
  boolean flag = true;
  long cnt = 0;
  for(int i = 0; i <al.size();  i++) {
  	
  	long mid = (al.get(i) + 1)/2;
  	if(bl.get(i)>=mid)
  	{
  	    flag = false;
  	    break;
  	}
  	else {
  		cnt++;
  	}
  	
  }
  
  
  if(flag) {
  	System.out.println(cnt);
  }
  else {
  	System.out.println(-1);
  }
  
  
  
  
  
  
  
  
  
  
  
  
  
  
    
    
  	
  		 
  	 
  	
       }
  
   
   }
  
  
  
  
  
  
  
  	
  
  
  
   
  
  }
   
  
   
   }
  
  
  
  
          
  
  
  
  
  		
  
  
          
  
  
  
  
  		