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("O3,unroll-loops")
  #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
  using namespace std;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <typename... Args>
  void logger(string vars, Args &&...values) {
    cerr << vars << " = ";
    string delim = "";
    (..., (cerr << delim << values, delim = ", "));
    cerr << endl;
  }
  const int MOD = 1e9 + 7;
  const int MX = 2e5 + 5;
  const long long int INF = 1e18;
  const long double PI = acos((long double)-1);
  const long double EPS = 1e-8;
  const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
  typedef array<long long int, 2> edge;
  long long int N;
  long long int get_next(long long int cur, vector<bool> &vis) {
    long long int ans = N + 1;
    for (long long int i = cur; i <= N; i++) {
      if (!vis[i]) {
        ans = i;
        break;
      }
    }
    return ans;
  }
  void solve() {
    cin >> N;
    vector<long long int> vec(N);
    for (long long int i = 0; i < (N); i++) cin >> vec[i];
    ;
    vector<bool> vis(N + 1, false);
    map<long long int, long long int> mp;
    for (long long int i : vec) {
      mp[i]++;
      if (i <= N) vis[i] = true;
    }
    vector<long long int> temp;
    for (auto pp : mp) {
      long long int num = pp.first, freq = pp.second;
      if (num <= N && vis[num]) freq--;
      for (long long int i = 0; i < freq; i++) temp.push_back(num);
    }
    if (temp.empty()) {
      cout << 0 << '\n';
      return;
    }
    sort(temp.begin(), temp.end());
    long long int vptr = 0;
    long long int ans = 0;
    for (long long int i : temp) {
      vptr = get_next(vptr + 1, vis);
      ans++;
      if (vptr >= (i + 1) / 2) {
        ans = -1;
        break;
      }
    }
    cout << ans << '\n';
  }
  clock_t startTime;
  double getCurrentTime() {
    return (double)(clock() - startTime) / CLOCKS_PER_SEC;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int T = 1;
    cin >> T;
    for (long long int _t = 1; _t <= T; _t++) {
      solve();
    }
    return 0;
  }
  