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 int row[4] = {1, 0, 0, -1};
  long long int col[4] = {0, -1, 1, 0};
  long long int mod = 1e9 + 7;
  bool is_prime(long long int n) {
    if (n % 2 == 0) return false;
    if (n % 3 == 0) return false;
    for (int i = 5; i * i <= n; i += 2) {
      if (n % i == 0) return false;
    }
    return true;
  }
  long long int power(long long int x, long long int y) {
    long long int temp;
    if (y == 0) return 1;
    temp = power(x, y / 2);
    if (y % 2 == 0)
      return temp * temp;
    else
      return x * temp * temp;
  }
  long long int BigMod(long long int a, long long int b, long long int m) {
    a %= m;
    long long int res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  long long int pankha(long long int n, long long int x) {
    if (n % x == 0)
      return (n / x);
    else
      return ((n / x) + 1);
  }
  void solve() {
    long long int n;
    cin >> n;
    vector<long long int> v, crr;
    map<long long int, long long int> mp;
    for (long long int i = 1; i <= n; i++) {
      long long int x;
      cin >> x;
      if (x <= n && !mp[x]) {
        v.push_back(x);
        mp[x]++;
      } else {
        crr.push_back(x);
      }
    }
    sort(crr.begin(), crr.end());
    bool flag = true;
    long long int cnt = 0;
    for (int i = n; i >= 1; i--) {
      if (!mp[i]) {
        long long int temp = crr.back();
        crr.pop_back();
        if (temp % (temp - i) != i) {
          flag = false;
          break;
        } else
          cnt++;
      }
    }
    if (flag)
      cout << cnt << "\n";
    else
      cout << -1 << "\n";
  }
  int main() {
    ios::sync_with_stdio(false);
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  