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 long long long long mod = 1e9 + 7;
  const long long long long mod2 = 998244353;
  long long long long power(long long long long a, long long long long b) {
    if (b == 0)
      return 1;
    else {
      long long long long x = power(a, b / 2) % mod;
      long long long long y = ((x % mod) * (x % mod)) % mod;
      if (b % 2) y = ((y % mod) * (a % mod)) % mod;
      return y % mod;
    }
  }
  long long long long log_a_to_base_b(long long long long a,
                                      long long long long b) {
    return log(a) / log(b);
  }
  bool compare_double(double x, double y) {
    double epsilon = 1e-6;
    if (fabs(x - y) < epsilon) return true;
    return false;
  }
  long long long long fact(long long long long n) {
    long long long long res = 1;
    for (long long long long i = 2; i <= n; i++) res = (res * i) % mod;
    return res;
  }
  long long long long nCr(long long long long n, long long long long r) {
    return (fact(n) / (fact(r) * fact(n - r)) % mod) % mod;
  }
  string dec_to_bin(long long n) {
    string ans = "";
    while (n != 0) {
      if (n % 2 == 0)
        ans += '0';
      else
        ans += '1';
      n = n / 2;
    }
    reverse(ans.begin(), ans.end());
    return ans;
  }
  bool sortbysec(const pair<long long, long long> &a,
                 const pair<long long, long long> &b) {
    return (a.second < b.second);
  }
  bool is_prime(long long n) {
    long long ans = 0;
    for (long long i = 1; i * i <= n; i++) {
      if (n % i == 0) {
        ans++;
        if (i != n / i) ans++;
      }
      if (ans > 2) return false;
    }
    return true;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL);
    long long tc;
    cin >> tc;
    while (tc--) {
      long long n;
      cin >> n;
      long long a[n];
      map<long long, long long> mp;
      vector<long long> vec;
      for (long long i = 0; i < n; i++) {
        cin >> a[i];
        if (mp[a[i]] == 0 && a[i] <= n)
          mp[a[i]]++;
        else
          vec.push_back(a[i]);
      }
      sort(vec.begin(), vec.end());
      long long ans = 0;
      long long j = 0;
      for (long long i = 1; i <= n; i++) {
        if (mp[i] == 1) continue;
        if (i <= (vec[j] - 1) / 2) {
          ans++;
          j++;
        } else {
          ans = -1;
          break;
        }
      }
      cout << ans << '\n';
    }
    return 0;
  }
  