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;
  void init() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> pos(n + 1, 0), a(n);
    for (long long i = 0; i < (n); i++) cin >> a[i];
    priority_queue<long long, vector<long long>, greater<long long>> pq;
    for (long long i = 0; i < (n); i++) {
      if (a[i] >= 1 and a[i] <= n and pos[a[i]] == 0)
        pos[a[i]] = 1;
      else
        pq.push(a[i]);
    }
    long long ans = 0;
    for (long long i = 1; i <= n; i++) {
      if (pos[i] == 1) continue;
      long long min_possible = pq.top();
      pq.pop();
      if (min_possible >= (2 * i + 1))
        ans++;
      else {
        cout << -1 << "\n";
        return;
      }
    }
    cout << ans << "\n";
  }
  signed main() {
    init();
    long long tt = 1;
    cin >> tt;
    while (tt--) {
      solve();
    }
    return 0;
  }
  