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 Q1617C {
  
  	public static void main(String[] args) {
  
  		Scanner scn = new Scanner(System.in);
          int t = scn.nextInt();
          while(t-->0){
              int n = scn.nextInt();
              int[] arr = new int[n];
              for(int i=0;i<n;i++){
                  int a = scn.nextInt();
                  arr[i] = a;
              }
              System.out.println(solve(arr));
          }
  
  	}
  	public static int solve(int[] arr) {
  		int n = arr.length;
  		int[] ans = new int[n];
          PriorityQueue<Integer> pq = new PriorityQueue<Integer>((a,b)->b-a);
          for(int i=0;i<n;i++) {
          	if(arr[i]<=n && ans[arr[i]-1]==0) 
          		ans[arr[i]-1] = arr[i];  
          	else
          		pq.add(arr[i]);
          }
          int an = pq.size();
          for(int i=n-1;i>=0;i--) {
          	if(ans[i]==0) {
          		int rem = pq.remove();
          		int pos = i+1;        		
          		if(rem%2==0) {
          			rem/=2;
          			rem--;
          		}else {
          			rem/=2;
          		}
          		if(rem<pos)
          			return -1;
          	}
          }
          return an;
  	}
  }
  