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.util.Arrays;
  import java.util.HashSet;
  import java.util.Set;
  import java.util.StringTokenizer;
  
  public class PaprikaAndPermutation {
      public static void main(String[] args) {
          
          FastReader fr = new FastReader();
          
          int ts = fr.nextInt();
          StringBuilder result = new StringBuilder();
  
          // for (int k = 1; k < 15; k++) {
  
          //     System.out.println("****");
          //     for (int j = k + 1; j >= 1; j--) {
  
                  // we know each number can have a resultant modulus from 0 to n / 2
                  // since from n / 2 to n - 1, the modulus is equal to n - k where k is between n / 2 to n - 1
  
          //     }
  
          // }
  
          for (int t = 0; t < ts; t++) {
  
              int n = fr.nextInt();
              Long[] arr = fr.nextLineAsLongArray(n);
  
              result.append(solve(arr) + "\n");
  
          }
  
          System.out.println(result.toString());
  
          // 3
          // 1 5 4
  
          // 5 -> 2 or 3
          // 5 mod 2 = 2 impossible to make into 3
          // 4 mod 
  
          // what are the conditions to make something into a remainder
  
          // remainder = x mod k
          // remainder can take values from 0 to min (k - 1, x)
          // 5 can become 5, 0, 1, 2, 1, 0
  
          // 4 mod 5 = 4
          // 4 mod 4 = 0
          // 4 mod 3 = 1
          // 4 mod 2 = 0
          // 4 mod 1 = 0
  
          // 5 mod 6 = 5
          // 5 mod 5 = 0
          // 5 mod 4 = 1
          // 5 mod 3 = 2
          // 5 mod 2 = 1
          // 5 mod 1 = 0
  
          // 6 mod 7 = 6
          // 6 mod 6 = 0
          // 6 mod 5 = 1
          // 6 mod 4 = 2
          // 6 mod 3 = 0
          // 6 mod 2 = 0
          // 6 mod 1 = 0
  
  
  
          // for each x, see what the maximum value we can make it,
          // set the item to the maximum
          // then, sort
          // see if there are any items s.t. it is less than the index
  
      }
  
      static long mod(long a, long b) {
          a = a % b;
          if (a < 0) {
              a += b;
          }
          return a;
      }
  
      static String solve(Long[] arr) {
  
          // 4 mod 1 = 0
          // 4 mod 2 = 0
          // 4 mod 3 = 1
          // 4 mod 4 = 0
  
          // 5 mod 1 = 0
          // 5 mod 2 = 1
          // 5 mod 3 = 2
          
          // 6 mod 1 = 0
          // 6 mod 2 = 0
          // 6 mod 3 = 0
          // 6 mod 4 = 2
          // 6 mod 5 = 1
          // 6 mod 6 = 0
  
          // if even, then can be between 0 to (n / 2) - 1
  
          Set<Long> seen = new HashSet<>();
          int count = 0;
  
          for (int i = 0; i < arr.length; i++) {
  
              if (seen.contains(arr[i]) || arr[i] > arr.length) {
  
                  if (arr[i] == 2) {
                      return "-1";
                  }
  
                  if (arr[i] % 2 == 0) {
                      arr[i] = - ((arr[i] / 2) - 1);
                  } else {
                      arr[i] = - (arr[i] / 2);
                  }
                  count++;
              } else {
  
                  seen.add(arr[i]);
  
              }
              
          }
  
          Arrays.sort(arr, (a, b) -> Long.compare(Math.abs(a), Math.abs(b)));
  
          for (int i = 0; i < arr.length; i++) {
  
              if (Math.abs(arr[i]) < i + 1) {
                  return "-1";
              }
  
          }
  
          return String.valueOf(count);
      }
  
  
      /** HELPER */
  
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
   
          public FastReader()
          {
              br = new BufferedReader(
                  new InputStreamReader(System.in));
          }
  
          int[] nextLineAsIntArray(int n) {
              int[] result = new int[n];
  
              for (int i = 0; i < n; i++) {
                  result[i] = nextInt();
              }
              return result;
          }
  
          Long[] nextLineAsLongArray(int n) {
              Long[] result = new Long[n];
  
              for (int i = 0; i < n; i++) {
                  result[i] = nextLong();
              }
  
              return result;
          }
   
          String next()
          {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
   
          int nextInt() { return Integer.parseInt(next()); }
   
          long nextLong() { return Long.parseLong(next()); }
   
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
   
          String nextLine()
          {
              String str = "";
              try {
                  str = br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }
  