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 read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 1) + (x << 3) + (ch ^ 48);
      ch = getchar();
    }
    return x * f;
  }
  inline void Prin(long long x) {
    if (x < 0) {
      putchar('-');
      x = -x;
    }
    if (x > 9) Prin(x / 10);
    putchar(x % 10 + '0');
  }
  const int qs = 1e5 + 7;
  int T, n;
  int a[qs], u[qs];
  int b[qs];
  int main() {
    T = read();
    while (T--) {
      n = read();
      int cnt = 0, flag = 1;
      for (int i = 1; i <= n; ++i) u[i] = 0;
      for (int i = 1; i <= n; ++i) a[i] = read();
      for (int i = 1; i <= n; ++i) {
        if (a[i] >= 1 && a[i] <= n && !u[a[i]]) {
          u[a[i]] = 1;
        } else
          b[++cnt] = a[i];
      }
      sort(b + 1, b + 1 + cnt);
      int j = 1;
      for (int i = 1; i <= n; ++i) {
        if (!u[i]) {
          int p = (b[j] + 1) / 2;
          j++;
          if (i >= p) {
            flag = 0;
          }
        }
      }
      if (!flag)
        cout << "-1\n";
      else
        cout << cnt << "\n";
    }
    return 0;
  }
  