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;
  template <class T>
  inline void read(T& x) {
    int f = 0;
    x = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
    while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
    x = f ? -x : x;
  }
  const int N = 1e5 + 10;
  const int INF = 0x7fffffff;
  const long long LINF = 0x3f3f3f3f3f3f3f3f;
  const double eps = 1e-8;
  const double PI = acos(-1);
  inline int tr(int x) { return (x - 1) >> 1; }
  bool vst[N];
  vector<int> a, b;
  int main() {
    int t;
    read(t);
    while (t--) {
      memset(vst, 0, sizeof(vst));
      a.clear();
      b.clear();
      int n, cur, ans = 0;
      read(n);
      for (int i = 1; i <= n; i++) {
        read(cur);
        if (cur > n || vst[cur])
          a.push_back(cur);
        else
          vst[cur] = true;
      }
      for (int i = 1; i <= n; i++)
        if (!vst[i]) b.push_back(i);
      sort(a.begin(), a.end());
      cur = a.size();
      for (int i = 0; i < cur; i++, ans++)
        if (b[i] > a[i] || b[i] > tr(a[i])) {
          ans = -1;
          break;
        }
      printf("%d\n", ans);
    }
    return 0;
  }
  