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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  const int N = 1e5 + 5;
  const long long int mod = 1e9 + 7;
  const long long int inf = 1e17;
  const long double PI = (acos(-1));
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      vector<int> arr(n);
      map<int, int> mp;
      for (int i = 0; i < n; ++i) {
        cin >> arr[i];
        ++mp[arr[i]];
      }
      bool flag = true;
      vector<int> res(n, 0);
      for (int i = 0; i < n; ++i) {
        if (mp.find(i + 1) != mp.end()) {
          --mp[i + 1];
          res[i] = i + 1;
        }
      }
      auto it = mp.begin();
      int ans = 0;
      for (int i = 0; i < n; ++i) {
        if (res[i] > 0) {
          continue;
        }
        while (it != mp.end() && it->second <= 0) {
          ++it;
        }
        if (it != mp.end() && (2 * (i + 1) < (it->first))) {
          ++ans;
          res[i] = i + 1;
          --(it->second);
        }
        if (res[i] == 0) {
          flag = false;
          break;
        }
      }
      if (!flag) {
        ans = -1;
      }
      cout << ans << endl;
    }
  }
  