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.HashMap;
  import java.util.HashSet;
  import java.util.Scanner;
  
  public class PaprikaandPermutation {
  	public static void solve(int[] arr,int n) {
  		HashSet<Integer> set=new HashSet<Integer>();
  		ArrayList<Integer> al=new ArrayList<Integer>();
  		for (int i = 0; i < arr.length; i++) {
  			if(arr[i]<=n) {
  				if(!set.contains(arr[i])) {
  					set.add(arr[i]);
  				}
  				else {
  					al.add(arr[i]);
  				}
  			}
  			else{
  				al.add(arr[i]);
  			}
  		}
  		Collections.sort(al,Collections.reverseOrder());
  		int count=al.size();
  		for (int i = 1; i <=n; i++) {
  			if(!set.contains(i)&&al.size() > 0) {
  				int x=al.remove(al.size()-1);
  				
  //				System.out.println(x%(x-i)+" "+i+" "+x);
  				if(x%(x-i)==i) {
  					
  					continue;
  				}
  				else {
  					System.out.println(-1);
  					return;
  				}
  			}
  		}
  		System.out.println(count);
  	}
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		Scanner in=new Scanner(System.in);
  		int t=in.nextInt();
  		for (int i = 0; i < t; i++) {
  			int n=in.nextInt();
  			int[] arr=new int[n];
  			for (int j = 0; j < n; j++) {
  				arr[j] = in.nextInt();
  			}
  			solve(arr, n);
  		}
  	}
  
  }
  