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.io.PrintWriter;
  import java.util.ArrayList;
  import java.util.Collections;
  import java.util.StringTokenizer;
  
  public class c {
  	public static void main(String[] args) {
  		FastScanner scan=new FastScanner();
  		PrintWriter out=new PrintWriter(System.out);
  		
  		int t=scan.nextInt();
  		t:for(int tt=0;tt<t;tt++) {
  			int n=scan.nextInt();
  			ArrayList<Integer> c=new ArrayList<>();
  			boolean[] v=new boolean[n+1];
  			for(int i=0;i<n;i++) {
  				int x=scan.nextInt();
  				if(x<=n&&!v[x]) {
  					v[x]=true;
  				}
  				else c.add(x);
  			}
  			Collections.sort(c);
  			int[] a=new int[c.size()];
  			for(int i=0;i<c.size();i++) a[i]=c.get(i);
  			
  			int ptr=n;
  			int res=0;
  			while(ptr>0&&v[ptr]) ptr--;
  			for(int i=a.length-1;i>=0;i--) {
  				if((a[i]+1)/2-1<ptr) {
  					out.println(-1);
  					continue t;
  				}
  				res++;
  				v[ptr]=true;
  				while(ptr>0&&v[ptr]) ptr--;
  			}
  			out.println(res);
  		}
  		
  		out.close();
  	}
  	static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastScanner() {
  			try	{
  				br = new BufferedReader(new InputStreamReader(System.in));
  				st = new StringTokenizer(br.readLine());
  			} catch (Exception e){e.printStackTrace();}
  		}
  
  		public String next() {
  			if (st.hasMoreTokens())	return st.nextToken();
  			try {st = new StringTokenizer(br.readLine());}
  			catch (Exception e) {e.printStackTrace();}
  			return st.nextToken();
  		}
  
  		public int nextInt() {return Integer.parseInt(next());}
  
  		public long nextLong() {return Long.parseLong(next());}
  
  		public double nextDouble() {return Double.parseDouble(next());}
  
  		public String nextLine() {
  			String line = "";
  			if(st.hasMoreTokens()) line = st.nextToken();
  			else try {return br.readLine();}catch(IOException e){e.printStackTrace();}
  			while(st.hasMoreTokens()) line += " "+st.nextToken();
  			return line;
  		}
  	}
  }