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.*;
  import java.io.*;
   
  public class C {
  	
  	static FastScanner fs = new FastScanner();
  	static PrintWriter pw = new PrintWriter(System.out);
  	static StringBuilder sb = new StringBuilder("");
  	
  	public static void main(String[] args) {
  		int t = fs.nextInt();
  		for (int tt = 0; tt < t; tt++) {
  			int n = fs.nextInt();
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++) {
  				a[i] = fs.nextInt();
  				a[i]--;
  			}
  			boolean[] used = new boolean[n];
  			ArrayList<Integer> b = new ArrayList<>();
  			for (int i = 0; i < n; i++) {
  				if (a[i] >= n) b.add(a[i]);
  				else if (!used[a[i]]) used[a[i]] = true;
  				else b.add(a[i]);
  			}
  			int m = b.size();
  			Collections.sort(b);
  			int cnt = 0;
  			boolean ok = true;
  			for (int i = 0; i < n; i++) {
  				if (!used[i]) {
  					if (i >= b.get(cnt) / 2) {
  						ok = false;
  					}
  				cnt++;
  				}
  			}
  			if(ok) sb.append(m + "\n");
  			else sb.append(-1 + "\n");
  		}
  		pw.print(sb.toString());
  		pw.close();
  	}
  	
  	static void sort(long[] a) {
  		ArrayList<Long> al = new ArrayList<>();
  		for (long i : a)
  			al.add(i);
  		Collections.sort(al);
  		for (int i = 0; i < a.length; i++)
  			a[i] = al.get(i);
  	}
   
  	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) {}return st.nextToken();}
  		int nextInt() {return Integer.parseInt(next());}
  		long nextLong() {return Long.parseLong(next());}
  		double nextDouble() {return Double.parseDouble(next());}
  		int[] readArray(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) {a[i] = nextInt();} return a;}
  	}
  }