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;
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      int arr[n], vis[200005] = {0};
      for (int i = 0; i < n; i++) {
        cin >> arr[i];
        if (arr[i] <= n) vis[arr[i]]++;
      }
      sort(arr, arr + n);
      vector<int> v;
      for (int i = 1; i <= n; i++) {
        if (!vis[i]) v.push_back(i);
      }
      int m = v.size(), l = 0;
      int ans = 0;
      for (int i = 0; i < n; i++) {
        int x = arr[i];
        if (x <= n && vis[x] > 1) {
          int r = vis[x], k = 1;
          while (r > 1) {
            if (l < m && (2 * v[l]) < x) {
              ans++;
              l++;
            } else {
              ans = -1;
              k = 0;
              break;
            }
            r--;
            i++;
          }
          if (!k) break;
        }
        if (x > n) {
          if (l < m && (2 * v[l]) < x) {
            ans++;
            l++;
          } else {
            ans = -1;
            break;
          }
        }
      }
      cout << ans << endl;
    }
    return 0;
  }
  