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: 
  def getint():
      return [int(i) for i in input().split()]
  def get():
      return int(input())
  def getstr():
      return [i for i in input().split()]
  def S():
      for test in range(int(input())):
          solve()
  import math
  import itertools as it
  import bisect
  import time
  import collections as ct
  
  def lower_bound(a,x):
      l=-1
      r=len(a)
      while l+1!=r:
          mid=l+r>>1
          if a[mid]<x:
              l=mid
          else:
              r=mid
      return r
  def upper_bound(a,x):
      l=-1
      r=len(a)
      while l+1!=r:
          mid=l+r>>1
          if a[mid]<=x:
              l=mid
          else:
              r=mid
      return r
  def solve():
      n=get()
      a=getint()
      a.sort()
      cha=list(range(1,n+1))
      cha=[0]+cha
      dic={}
      for i in range(n):
          if 1<=a[i]<=n and dic.get(a[i],0)!=1:
              cha[a[i]]=0
              dic[a[i]]=1
              a[i]=0
      # print(a)
      # print(cha)
      check=[]
      for i in range(n):
          if a[i]!=0:
              check.append(a[i])
      
      C=[]
      for i in cha:
          if i!=0:
              C.append(i)
      # print(C)
      for i in range(len(check)):
          if check[i]&1:
              check[i]//=2
          else:
              check[i]=check[i]//2-1
      # print(check)
      if len(check)!=len(C):
          print(-1)
          return
      check.sort()
      C.sort()
      for i in range(len(check)):
          if check[i]<C[i]:
              print(-1)
              return
      print(len(check))
  S()
  