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: # cook your dish here
  import sys
  #import random
  from bisect import bisect_left as lb
  from bisect import bisect_right as rb
  from collections import deque
  #sys.setrecursionlimit(10**8)
  from queue import PriorityQueue as pq
  from math import gcd
  #import math
  input_ = lambda: sys.stdin.readline().strip("\r\n")
  ii = lambda : int(input_())
  il = lambda : list(map(int, input_().split()))
  ilf = lambda : list(map(float, input_().split()))
  lii = lambda : list(map(int, list(ip())))
  ip = lambda : input_()
  fi = lambda : float(input_())
  ap = lambda ab,bc,cd : ab[bc].append(cd)
  li = lambda : list(input_())
  pr = lambda x : print(x)
  prinT = lambda x : print(x)
  f = lambda : sys.stdout.flush()
  inv =lambda x:pow(x,mod-2,mod)
  dx = [0,0,1,-1]
  dy = [1,-1,0,0]
  mod = 10**9 + 7
  mod1 = 998244353
  
  
  for _ in range (ii()) :
      n = ii()
      a = il()
  
      a.sort()
  
      b = [0 for i in range (n+1)]
  
      c = []
  
      for i in range (n) :
          if (a[i] <= n) :
              b[a[i]] += 1
  
      for i in range (1,1+n) :
          if (b[i] == 0) :
              c.append(i)
  
      ans = len(c)
  
      for i in range (n-1,-1,-1) :
          if (len(c) == 0) :
              break
  
          fl = 1
  
          if (a[i] <= n and b[a[i]] == 1) :
              fl = 0
  
          if (fl and a[i]>2*c[-1] and c[-1] != 1) :
              if (a[i] <= n) :
                  b[a[i]] -= 1
  
              a[i] = c[-1]
              b[c[-1]] += 1
              c.pop()
  
          if (len(c) > 0 and c[-1] == 1) :
              ind = 0
  
              for i in range (n) :
                  if (a[i] > n) :
                      ind = a[i]
                      break
                  elif (b[a[i]] > 1) :
                      ind = a[i]
                      break
  
              if (ind != 2) :
                  c.pop()
  
  
      if (len(c) == 0) :
          print(ans)
      else :
          print(-1)
  
          
  
      
      
  