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>
  using namespace std;
  int a[100100];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ;
    int tt;
    cin >> tt;
    while (tt--) {
      long long n;
      cin >> n;
      multiset<long long> mls;
      for (int i = 0; i < n; ++i) {
        cin >> a[i];
        mls.insert(a[i]);
      }
      long long ans = 0;
      for (int i = n; i > 0; --i) {
        if (mls.find(i) != mls.end()) {
          mls.erase(mls.find(i));
        } else {
          ans++;
          long long x = *(--mls.end());
          mls.erase(mls.find(x));
          if (2 * i >= x) {
            ans = -1;
            break;
          }
        }
      }
      cout << ans << '\n';
    }
  }
  