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;
  const int MAXN = 2e5 + 55;
  long long a[MAXN], mark[MAXN];
  vector<long long> ans;
  int main() {
    long long T, T1;
    cin >> T;
    for (T1 = 1; T1 <= T; T1++) {
      long long n, i, j, sz, st, vall, f2 = 0, lg = 0;
      cin >> n;
      for (i = 1; i <= n; i++) cin >> a[i];
      sort(a + 1, a + n + 1);
      for (i = 1; i <= n; i++) {
        if (a[i] > n || (a[i] <= n && mark[a[i]])) {
          ans.push_back(a[i]);
          continue;
        }
        mark[a[i]] = 1;
      }
      sz = ans.size();
      st = 0;
      for (i = 1; i <= n; i++) {
        if (mark[i] == 0) {
          vall = ans[st];
          if (((i * 2) + 1) > vall) {
            f2 = 1;
            break;
          }
          st++;
          lg++;
        }
      }
      if (f2)
        cout << -1 << endl;
      else
        cout << lg << endl;
      for (i = 1; i <= n; i++) mark[i] = 0;
      ans.clear();
    }
  }
  