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.HashMap;
  import java.util.Arrays;
  import java.util.Scanner;
  public class C1617{
      
   public static int[] sort(int[] arr,int low,int high){
          int[] sortedArray;
          if (low==high) {
              sortedArray=new int[]{arr[low]};
              return sortedArray;
          }
          int mid=(low+high)/2;
          int[] arr1=sort(arr,low,mid);
          int[] arr2=sort(arr,mid+1,high);
  
          sortedArray=merge(arr1,low,mid,arr2,mid+1,high);
          return sortedArray;
  
      }
  
      public static int[] merge(int[] arr1,int low1,int high1,int[] arr2,int low2,int high2){
  
          int size1=high1-low1+1;
          int size2=high2-low2+1;
          
          int[] arr=new int[size1+size2];
          int pointer=0;
          low1=0;low2=0;
          while (low1<size1 && low2<size2){
              if (arr1[low1] <= arr2[low2]) {
                  arr[pointer++]=arr1[low1++];
              }
              else{
                  arr[pointer++]=arr2[low2++];
              }
  
          }
  
          while(low1<size1) arr[pointer++]=arr1[low1++];
          while(low2<size2) arr[pointer++]=arr2[low2++];
          return arr;
      }
      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];
                  for(int i=0;i<n;i++) arr[i]=sc.nextInt();
                  arr=sort(arr,0,n-1);
                  int next=1;
                  int ans=0;
                  int f[]=new int[n+1];
                  for(int i=0;i<n;i++){
                      if (arr[i]==next) {
                          f[next]=1;
                          while(next <=n && f[next]==1) next++;
                      }
                      else{
                          if (arr[i] <= n && f[arr[i]]==0) {
                              f[arr[i]]=1;
                          }
                          else{
                              if (next < (arr[i]+1)/2) {
                                  ans++;
                                  f[next]=1;
                                  while(next <=n && f[next]==1) next++;
                              }
                              else {
                                  ans=-1;
                                  break;
                              }
                          }
                      }
                  }        
                  System.out.println(ans);
          }
          sc.close();
      }   
  }
  