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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  template <typename T>
  long long powe(T base, T to) {
    long long ans = 1;
    while (to > 0) {
      if (to & 1)
        ans = ((ans) * (base)), to--;
      else
        base = ((base) * (base)), to /= 2;
    }
    return ans;
  }
  template <typename T>
  T gcd(T a, T b) {
    return (b == 0) ? a : gcd(b, a % b);
  }
  bool prime(long long n) {
    if (n == 1) return 0;
    for (long long i = 2; i * i <= n; i++)
      if (n % i == 0) return 0;
    return 1;
  }
  int binarySearch(int arr[], int start, int end, int val) {
    int mid;
    end--;
    while (start <= end) {
      mid = start + (end - start) / 2;
      if (arr[mid] == val)
        return mid;
      else if (arr[mid] < val)
        start = mid + 1;
      else
        end = mid - 1;
    }
    return -1;
  }
  long long fun(int arr[], int n, int x) {
    long long ans = x;
    for (int i = 0; i < n - 1; i++) ans += min(x, arr[i + 1] - arr[i]);
    return ans;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int tl = 1;
    cin >> tl;
    while (tl--) {
      int n;
      cin >> n;
      int arr[n];
      for (auto &i : arr) cin >> i;
      ;
      sort(arr, arr + n);
      bool ok = true;
      for (int i = 0; i < n; i++)
        if (arr[i] < (i + 1)) {
          ok = false;
          break;
        }
      if (!ok) {
        cout << "-1\n";
        continue;
      }
      vector<bool> tmp(n, 0);
      int ans = n;
      int val = 1;
      for (int i = 0; i < n; i++) {
        if (arr[i] >= 1 && arr[i] <= n && !tmp[arr[i] - 1]) {
          tmp[arr[i] - 1] = 1, ans--;
          while (tmp[val - 1] && val <= n) val++;
          continue;
        }
        if (arr[i] == val) {
          tmp[val - 1] = 1;
          while (tmp[val - 1] && val <= n) val++;
          continue;
        }
        if ((arr[i] - val) <= val) {
          ans = -1;
          break;
        }
        tmp[val - 1] = 1;
        while (tmp[val - 1] && val <= n) val++;
      }
      cout << ans << "\n";
    }
  }
  