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
  BUFSIZE = 8192
  class FastIO(IOBase):
      newlines = 0
      def __init__(self, file):
          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 = os.read(self._fd, max(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 = os.read(self._fd, max(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: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)
  try:sys.stdin,sys.stdout=open('in.txt','r'),open('out.txt','w')
  except:pass
  ii1=lambda:int(sys.stdin.readline().strip()) # for interger
  is1=lambda:sys.stdin.readline().strip() # for str
  iia=lambda:list(map(int,sys.stdin.readline().strip().split())) # for List[int]
  isa=lambda:sys.stdin.readline().strip().split() # for List[str]
  mod=int(1e9 + 7);
  from math import inf 
  from collections import defaultdict as dd
  # from math import *
  # from collections import *;
  # from collections import deque as dq 
  # from string import ascii_lowercase,ascii_uppercase
  # from functools import lru_cache
  # sys.setrecursionlimit(500000)
  # from bisect import bisect_left as bl,bisect_right as br
  from heapq import heapify,heappop,heappush
  # def google(p):print('Case #'+str(p)+': ',end='')
  ###################### Start Here ######################
  from math import gcd
  def main(tt):
      n = ii1()
      arr = iia()
      arr.sort()
      vis = [0]*(n+1)
      heap = []
      ans = 0 
      for i in range(n-1,-1,-1):
          val = arr[i]
          if val <= n and vis[val]==0:
              vis[val] = 1 
              continue
          m = arr[i]//2 
          r = val % (m+1)
          ans += 1
          heappush(heap,-r)
      for i in range(n,0,-1):
          if vis[i]==0:
              if heap:
                  val = - heappop(heap)
                  if val < i:
                      print(-1);return 
              else:
                  print(-1)
                  return 
      print(ans)
  
  
  
  
  if __name__ == "__main__":
      t = 1 
      t = ii1()
      for _ in range(t):
          main(_)
      
  # snippets
  # @f - for loop 
  # @bfs - breadth first search 
  # @Trie - Trie Data Structure 
  # @rangeupdatepointquery - range update
  # @DS - Disjoint Set
  # @seive - generate seive