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;
  template <class T>
  void _print(T t) {
    cerr << t;
  }
  template <class T>
  void _print(array<T, 2> p) {
    cerr << "(";
    _print(p[0]);
    cerr << ",";
    _print(p[1]);
    cerr << ")";
  }
  template <class T>
  void _print(vector<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) {
      cerr << "{";
      _print(i.first);
      cerr << ":";
      _print(i.second);
      cerr << "} ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "{ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "}";
  }
  template <class T>
  void _print(unordered_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 << "}";
  }
  const long long MOD = 1e9 + 7;
  long long pow(long long x, long long y) {
    long long res = 1;
    while (y > 0) {
      if (y & 1) (res *= x) %= MOD;
      (x *= x) %= MOD, y >>= 1;
    }
    return res;
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> A(n);
    vector<int> B(n + 1, 0);
    vector<int> C, D;
    for (int i = 0; i < n; i++) {
      cin >> A[i];
      if (A[i] > n) {
        C.push_back(A[i]);
      } else if (B[A[i]]) {
        C.push_back(A[i]);
      } else {
        B[A[i]] = 1;
      }
    }
    for (int i = 1; i <= n; i++) {
      if (B[i]) continue;
      D.push_back(i);
    }
    sort(C.begin(), C.end());
    int m = (int)C.size();
    for (int i = 0; i < m; i++) {
      int k = C[i] - D[i];
      if (C[i] % k != D[i]) {
        cout << "-1\n";
        return;
      }
    }
    cout << m << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    cin >> t;
    for (int tc = 1; tc <= t; tc++) {
      solve();
    }
    return 0;
  }
  