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: from sys import stdin,stdout,setrecursionlimit
  from math import gcd,sqrt,factorial,pi,inf
  from collections import deque,defaultdict
  from bisect import bisect,bisect_left
  from time import time
  from itertools import permutations as per
  from heapq import heapify,heappush,heappop,heappushpop
  input=stdin.readline
  R=lambda:map(int,input().split())
  I=lambda:int(input())
  S=lambda:input().rstrip('\r\n')
  L=lambda:list(R())
  P=lambda x:stdout.write(str(x)+'\n')
  lcm=lambda x,y:(x*y)//gcd(x,y)
  nCr=lambda x,y:(f[x]*inv((f[y]*f[x-y])%N))%N
  inv=lambda x:pow(x,N-2,N)
  sumx=lambda x:(x**2+x)//2
  N=10**9+7
  
  for _ in range(I()):
      n=I()
      a=L()
      e=[]
      p=[0]*n
      for i in a:
          if i<=n and not p[i-1]:
              p[i-1]=i
          else:
              e+=i,
      e.sort(reverse=True)
      ans=0
      for i in range(n):
          if not p[i]:
              if (e.pop()+1)//2-1<i+1:
                  ans=-1
                  break
              ans+=1
      print(ans)