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 sys
  import os.path
   
  if(os.path.exists('input.txt')) :
      sys.stdin = open("input.txt", "r")
      sys.stdout = open("output.txt", "w")
      sys.stderr = open("error.txt", "w")
      
  depth = 200001
  mod = 1000000007 
  lim = mod * mod
  sys.setrecursionlimit(depth) 
   
  linp = lambda: list(minp())
  minp = lambda: map(int, input().split())
   
  from math import inf, ceil, sqrt, log2, gcd
  from collections import defaultdict, deque
  
  dxy = [(1, 0),(-1, 0),(0, 1),(0, -1)]
  
  def expo(a, b) :
      if not b : return 1
      v = expo(a, b//2)
      v = (v * v) % mod
      if b%2 : v = (v * a) % mod
      return v
  
  def modinv(a, b) :
      d = expo(b, mod-2)
      v = (a * d) % mod
      return v
  
  def checkPrime(n) :
      if n < 4 : return True
      if n%2 == 0 : return False
      for i in range(3, int(sqrt(n))+1, 2) :
          if n%i == 0 :
              return False
      return True    
  
  def highestPowerof2(x):
      x |= x >> 1
      x |= x >> 2
      x |= x >> 4
      x |= x >> 8
      x |= x >> 16
  
      return x ^ (x >> 1)
  
  for _ in range(int(input())) :
      n = int(input())
      a = sorted(linp())
      b = [False for _ in range(n)]
      c = [True for _ in range(n)]
  
      for i in range(n) :
          if a[i] <= n and not b[a[i]-1] :
              c[i] = False 
              b[a[i]-1] = True 
  
      d = list()
      for i in range(n) :
          if c[i] :
              d.append(a[i])
  
      e = list()
      for i in range(n) :
          if not b[i] :
              e.append(i+1)
  
      a, b = d, e
      a.sort(), b.sort()
      
      n, f = len(a), True
      for i in range(n) :
          if b[i] >= (a[i]-b[i]) :
              f = False
  
      if f : print(n)
      else : print(-1)