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 read(vector<T> &a) {
    for (auto &i : a) cin >> i;
  }
  template <class T>
  void read1(vector<T> &a, int n) {
    for (int i = 1; i <= n; i++) cin >> a[i];
  }
  template <class T>
  void _print(T x) {
    cerr << x << endl;
  }
  template <class S, class T>
  void _print(pair<S, T> x) {
    cerr << "{" << x.first << "," << x.second << "}" << endl;
  }
  template <class T>
  void _print(vector<T> x) {
    cerr << "[";
    for (auto i : x) cerr << i << ", ";
    cerr << "]\n";
  }
  template <class S, class T>
  void _print(vector<pair<S, T>> x) {
    for (auto i : x) cerr << "[" << i.first << "," << i.second << "]" << endl;
  }
  template <class T>
  void _print(set<T> x) {
    cerr << "{ ";
    for (auto i : x) cerr << i << ", ";
    cerr << "}\n";
  }
  template <class S, class T>
  void _print(map<S, T> x) {
    for (auto i : x) {
      cerr << i.first << ": " << i.second << endl;
    }
  }
  long long exp(int a, int b, int p) {
    long long res = 1;
    a = a % p;
    while (b) {
      if (b & 1) res = (res * a) % p;
      a = ((long long)a * a) % p;
      b = b >> 1;
    }
    return res;
  }
  long long mmiPrime(int a, int p) { return exp(a, p - 2, p); }
  pair<int, int> extendedGCD(int a, int b) {
    if (b == 0) {
      pair<int, int> ans;
      ans.first = 1;
      ans.second = 0;
      return ans;
    }
    pair<int, int> temp = extendedGCD(b, a % b);
    pair<int, int> ans;
    ans.first = temp.second;
    ans.second = temp.first - (a / b) * temp.second;
    return ans;
  }
  int mmi(int a, int m) {
    pair<int, int> tmp = extendedGCD(a, m);
    return tmp.first;
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n + 1, 0);
    vector<int> bad;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      if (x >= 1 && x <= n && a[x] != 1)
        a[x] = 1;
      else
        bad.push_back(x);
    }
    int ans = 0;
    sort(bad.begin(), bad.end());
    int j = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] != 1) {
        if (j >= bad.size()) {
          cout << -1 << endl;
          return;
        }
        int x = bad[j];
        int rem = x / 2;
        if (x % 2 == 0) rem--;
        if (i > rem) {
          cout << -1 << endl;
          return;
        } else {
          a[i] = 1;
          j++;
          ans++;
        }
      }
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    clock_t time = clock();
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    time = clock() - time;
    return 0;
  }
  