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 pap {
  	public static void main(String args[]) {
  	   Scanner sc=new Scanner(System.in);
  	   int t=sc.nextInt();
  	   while(t-->0) {
  		   int n=sc.nextInt();
  		   int arr[]=new int[n];
  		   ArrayList<Integer> a=new ArrayList<>();
  		   for(int i=0;i<n;i++) {
  			   arr[i]=sc.nextInt();
  		   }
  		   for(int i=0;i<n;i++) {
  			   int u=Math.abs(arr[i]);
  			   if(u<=n) {
  				   if(arr[u-1]>0) {
  					   arr[u-1]=-arr[u-1];
  				   }
  				   else a.add(u); 
  			   }else a.add(u);
  		   }
  		   Collections.sort(a);
  		   int i=0;
  		   boolean flag=false;
  		   for(int j=0;j<a.size();j++) {
  			   while(i<n && arr[i]<0) i++;
  			   if(i<n) {
  				   if(i+1<=((a.get(j)+1)/2)-1) {
  					   i++;
  					   while(i<n && arr[i]<0) i++;
  				   }
  				   else {
  					   flag=true;
  					   break;
  				   }
  			   }
  		   }
  		   while(i<n && arr[i]<0) i++;
  		   if(flag || i<n) System.out.println(-1);
  		   else System.out.println(a.size());
  		   
  		  
  		   
  	   }
  	   sc.close();
  }
  
  	}