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 collections import *
  import sys
  import io, os
  import math
  from heapq import *
  gcd = math.gcd
  sqrt = math.sqrt
  def ceil(a, b):
      a = -a
      k = a // b
      k = -k
      return k
  # arr=list(map(int, input().split()))
  input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  def strinp(testcases):
      k = 5
      if (testcases == -1 or testcases == 1):
          k = 1
      f = str(input())
      f = f[2:len(f) - k]
      return f
  
  def main():
      t=int(input())
      for _ in range(t):
          n=int(input())
          arr=list(map(int, input().split()))
          perm=[0]*(n+1)
          perm[0]=1
          useful=[]
          for i in range(n):
              if(arr[i]>n):
                  useful.append(arr[i])
                  continue
              if(perm[arr[i]]!=0):
                  useful.append(arr[i])
                  continue
              perm[arr[i]]=1
          c=0
          if(len(useful)==0):
              print(0)
              continue
          useful.sort()
          pos=[]
          for i in range(1,n+1):
              if(perm[i]==0):
                  pos.append(i)
          for j in range(len(useful)):
              if(pos[j]>=useful[j]-pos[j]):
                  c=-1
                  break
              c+=1
          print(c)
  main()