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 mod = 1e9 + 7;
  const int MOD = 998244353;
  const long long int maxn = 1e5 + 5;
  const long long int inf = 1e12;
  const double pi = acos(-1);
  inline long long int add(long long int x, long long int y) {
    x += y;
    if (x >= mod) x -= mod;
    return x;
  }
  inline long long int sub(long long int x, long long int y) {
    x -= y;
    if (x < 0) x += mod;
    return x;
  }
  inline long long int mul(long long int x, long long int y) {
    return (x * 1ll * y) % mod;
  }
  void _print(long long int t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long int t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.ff);
    cerr << ",";
    _print(p.ss);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  long long int n, m, x, y, z, u, minn, maxx;
  void solve() {
    cin >> n;
    vector<long long int> arr(n);
    vector<long long int> rem;
    set<long long int> st;
    for (long long int i = 1; i <= n; i++) st.insert(i);
    for (long long int i = 0; i < n; i++) {
      cin >> arr[i];
      if (st.find(arr[i]) != st.end()) {
        st.erase(arr[i]);
      } else {
        rem.push_back(arr[i]);
      }
    }
    sort(rem.begin(), rem.end());
    for (int i = 0; i < rem.size(); i++) {
      long long int curr = *(st.begin());
      if (rem[i] > 2 * curr) {
        st.erase(st.begin());
      } else {
        cout << -1 << endl;
        return;
      }
    }
    cout << rem.size() << endl;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie();
    cout.tie();
    long long int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  