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: #include <bits/stdc++.h>
  int main() {
    int tasks;
    std::cin >> tasks;
    for (int o = 0; o < tasks; ++o) {
      int n;
      std::cin >> n;
      int res = 0;
      std::vector<int> a(n);
      std::vector<int> b(n);
      std::vector<int> need_change;
      for (int i = 0; i < n; ++i) {
        std::cin >> a[i];
        if (a[i] <= n && b[a[i] - 1] == 0) {
          b[a[i] - 1] = 1;
        } else {
          need_change.push_back(a[i]);
        }
      }
      std::vector<int> change_to;
      for (int i = 0; i < n; ++i) {
        if (b[i] == 0) {
          change_to.push_back(i + 1);
        }
      }
      std::sort(need_change.begin(), need_change.end());
      bool can_we = 1;
      for (int i = 0; i < need_change.size(); ++i) {
        int max_get_from_op = need_change[i] / 2;
        if (need_change[i] % 2 == 1) {
          max_get_from_op += 1;
        }
        if (change_to[i] >= max_get_from_op) {
          can_we = 0;
        }
      }
      if (can_we) {
        std::cout << need_change.size() << std::endl;
      } else {
        std::cout << -1 << std::endl;
      }
    }
  }
  