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;
  long long int power(long long int a, long long int b) {
    long long int res = 1;
    a = a % 1000000007;
    while (b > 0) {
      if (b & 1) {
        res = (res * a) % 1000000007;
        b--;
      }
      a = (a * a) % 1000000007;
      b >>= 1;
    }
    return res;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      long long int n, match = 0;
      cin >> n;
      vector<long long int> a(n), v(n + 1);
      map<long long int, long long int> m;
      for (int i = 0; i < n; ++i) {
        cin >> a[i];
        if (a[i] <= n && v[a[i]] == 0) match++, v[a[i]]++;
      }
      sort(a.begin(), a.end());
      for (int i = 1; i <= n; ++i) {
        if (binary_search(a.begin(), a.end(), i)) {
          m[upper_bound(a.begin(), a.end(), i) - a.begin() - 1] = 1;
        }
      }
      long long int ptr = n, i = n - 1, c = 0;
      while (ptr && i >= 0) {
        if (v[ptr])
          ptr--;
        else {
          while (m[i]) i--;
          if (a[i] <= 2 * ptr) {
            cout << -1 << '\n';
            c++;
            break;
          } else {
            v[ptr] = 1;
            ptr--;
            m[i] = 1;
          }
        }
      }
      if (c == 0) {
        cout << n - match << '\n';
      }
    }
  }
  