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 pc;
  void PrintCase() { printf("Case #%lli: ", ++pc); }
  long long int bigmod(long long int n, long long int p) {
    if (p == 0) return 1;
    if (p == 1) return (n + 1000000007) % 1000000007;
    if (p % 2)
      return (bigmod(n, p - 1) * n + 1000000007) % 1000000007;
    else {
      long long int x = bigmod(n, p / 2);
      return (x * x + 1000000007) % 1000000007;
    }
  }
  long long int power(long long int n, long long int p) {
    if (p == 0) return 1;
    if (p == 1) return n;
    if (p % 2)
      return power(n, p - 1) * n;
    else {
      long long int x = power(n, p / 2);
      return x * x;
    }
  }
  int main() {
    long long int t;
    cin >> t;
    while (t--) {
      long long int n, i, j, k;
      cin >> n;
      long long int a[n + 5], vis[n + 5];
      memset(vis, 0, sizeof(vis));
      vector<long long int> v;
      for (i = 0; i < n; i++) {
        cin >> a[i];
        if (a[i] <= n and vis[a[i]] == 0) {
          vis[a[i]] = 1;
        } else
          v.push_back(a[i]);
      }
      sort(v.begin(), v.end());
      long long int ok = 0;
      for (i = 1, j = 0; i <= n and j < v.size(); i++) {
        if (vis[i]) continue;
        long long int x = v[j];
        if (x % (x - i) == i) {
          vis[i] = 1;
          ok++;
        } else {
          ok = -1;
          break;
        }
        j++;
      }
      cout << ok << endl;
    }
  }
  