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;
  vector<int> v1[200005];
  bool visit[200005];
  bool color[200005];
  int parent[200005];
  int zz, zz1;
  vector<int> vv;
  void solve() {
    long long int n, m;
    cin >> n;
    vector<long long int> v1(n), store;
    set<long long int> set1;
    for (long long int x = 1; x < n + 1; ++x) set1.insert(x);
    for (long long int x = 0; x < n; ++x) {
      cin >> m;
      if (set1.count(m) > 0)
        set1.erase(m);
      else
        store.push_back(m);
    }
    sort(store.begin(), store.end());
    long long int count = 0, f1 = 1;
    for (long long int x = 0; x < store.size(); ++x) {
      long long int a = store[x];
      auto it = set1.begin();
      long long int pehla_ele = *it;
      long long int garaj = (a - 1) / 2;
      if (pehla_ele > garaj) {
        cout << -1 << endl;
        return;
      } else {
        set1.erase(pehla_ele);
        count++;
      }
    }
    cout << count << endl;
  }
  void test_case() {
    long long int t;
    cin >> t;
    while (t-- > 0) {
      solve();
    }
  }
  int main() { test_case(); }
  