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.*;
  import java.util.*;
  
  public class C {
  	
  	public static void main(String[] args)throws IOException {
  		
  		FastScanner scan = new FastScanner();
  		PrintWriter output = new PrintWriter(System.out);
  		int t = scan.nextInt();
  		for(int tt = 0;tt<t;tt++) {
  			int n = scan.nextInt();
  			int arr[] = scan.readArray(n);
  			HashSet<Integer> seen = new HashSet<>();
  			for(int i = 0;i<n;i++) {
  				if(arr[i]<=n && !seen.contains(arr[i])) {
  					seen.add(arr[i]);
  					arr[i] = 0;
  				}
  			}
  			sort(arr);
  			int notSeen = 1;
  			boolean possible = true;
  			for(int i = 0;i<n;i++) {
  				if(arr[i]==0) continue;
  				if(seen.contains(notSeen)) {
  					notSeen++;
  					i--;
  					continue;
  				}
  				if((arr[i]-1)/2 < notSeen) {
  					possible = false;
  					break;
  				}
  				notSeen++;
  			}
  			output.println(possible ? n-seen.size() : -1);
  			
  		}
  		output.flush();
  	}
  	
  	public static int[] sort(int arr[]) {
  		List<Integer> list = new ArrayList<>();
  		for(int i:arr) list.add(i);
  		Collections.sort(list);
  		for(int i = 0;i<list.size();i++) arr[i] = list.get(i);
  		return arr;
  	}
  	
  	public static long gcd(long a, long b) {
  		if(a == 0) return b;
  		return gcd(b%a, a);
  	}
  	
  	public static void printArray(int arr[]) {
  		for(int i:arr) System.out.print(i+" ");
  		System.out.println();
  	}
  	
  	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());
  		}
  		
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  
  }
  