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 os, sys
  from io import BytesIO, IOBase
  from types import GeneratorType
  from bisect import *
  from collections import defaultdict, deque, Counter
  import math, string
  from heapq import *
  from operator import add
  from itertools import accumulate
  
  BUFSIZE = 8192
  sys.setrecursionlimit(10 ** 5)
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          import os
  
          self.os = os
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              self.os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  
  inf = float("inf")
  en = lambda x: list(enumerate(x))
  ceil_ = lambda a, b: (a + b - 1) // b
  
  
  ii = lambda: int(input())
  r = lambda: map(int, input().split())
  rr = lambda: list(r())
  
  
  # --------------------------
  
  def ok(a,b):
      d = a - b
      if d < 1:
          return False
  
      return a % d == b
  
  
  def solve():
      n = ii()
      arr = rr()
      arr.sort(reverse=False)
      brr = Counter(arr)
  
      done = defaultdict(bool)
      for i in arr:
          if i <= n and not done[i]:
              done[i] = True
              brr[i] -= 1
  
      crr = []
      for i,j in brr.items():
          crr.extend([i] * j)
      
      # print(crr)
      crr.sort()
      j = c = 0
      for i in range(1, n+ 1):
          if not done[i]:
              if crr[j] <= 2 * i:
                  print(-1)
                  return
              else:
                  done[i] = True
                  c += 1
                  j += 1
  
      print(c)
  
  
  
  
  for _ in " " * ii():
      solve()