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;
  long long mod(long long x, long long MOD) {
    if (x > MOD) x -= MOD;
    return x;
  }
  long long _ceil(long long x, long long y) { return x / y + (x % y != 0); }
  void wrt() {
    cout << "\n";
    return;
  }
  template <typename T1, typename... T2>
  void wrt(T1 x, T2... args) {
    ((cout << x << ' '), wrt(args...));
  }
  template <typename T>
  ostream& operator<<(ostream& os, const vector<T>& v) {
    for (auto& i : v) os << i << " ";
    return os;
  }
  template <typename T>
  istream& operator>>(istream& in, vector<T>& v) {
    for (auto& e : v) in >> e;
    return in;
  }
  const long long MOD = 1e9 + 7;
  const long long N = 1e6 + 5, inf = 1e18 + 2;
  long long n, m, x, u, v, k, p, q, i, j;
  string s, t;
  long long ans;
  long long a[N];
  void solve() {
    cin >> n;
    vector<long long> a(n);
    cin >> a;
    multiset<long long> st((a).begin(), (a).end());
    x = 0;
    vector<long long> vis(n + 1, 0);
    for (long long i = 1; i <= (n); ++i) {
      if (st.find(i) != st.end()) {
        st.erase(st.find(i));
        vis[i] = 1;
      }
    }
    for (long long i = 1; i <= (n); ++i) {
      if (!vis[i]) {
        if (*st.begin() - i > i) {
          vis[i] = 1;
          st.erase(st.begin());
        }
        x++;
      }
    }
    for (long long i = 1; i <= (n); ++i)
      if (!vis[i]) return wrt(-1);
    wrt(x);
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  