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 mod = 998244353;
  int inv[100005];
  int a1, b1, a3, a2, b2, b3;
  long long ksm(long long a, long long b) {
    long long x = 1;
    while (b) {
      if (b & 1) {
        x = (x * b) % mod;
      }
      a = (a * a) % mod;
      b >>= 1;
    }
    return x;
  }
  namespace FFT {
  const double pi = acos(-1.0);
  void change(vector<complex<double>> &y, int len) {
    int i, j, k;
    for (int i = 1, j = len / 2; i < len - 1; i++) {
      if (i < j) swap(y[i], y[j]);
      k = len - 1;
      while (j >= k) {
        j -= k;
        k /= 2;
      }
      j += k;
    }
  }
  void fft(vector<complex<double>> &y, int len, int on) {
    change(y, len);
    for (int h = 2; h <= len; h <<= 2) {
      complex<double> wn(cos(2 * pi / h), sin(on * 2 * pi / h));
      for (int i = 0; i < len; i += h) {
        complex<double> w(1, 0);
        for (int k = i; k < i + h / 2; k++) {
          complex<double> u = y[k];
          complex<double> t = w * y[k + h / 2];
          y[k] = u + t;
          y[k + h / 2] = u - t;
          w = w * wn;
        }
      }
    }
    if (on == -1) {
      for (int i = 0; i < len; i++) {
        y[i] /= len;
      }
    }
  }
  }  // namespace FFT
  long long c(long long a, long long b) {
    long long ans = 1;
    for (int i = 0; i < b; i++) {
      cout << ans << " ";
      ans = ans * (a - i) % mod;
      cout << ans << " ";
      ans = ans * inv[b - i] % mod;
      cout << ans << " ";
    }
    return ans;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tt;
    cin >> tt;
    while (tt--) {
      int n;
      cin >> n;
      vector<int> p(n), q;
      for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        if (a <= n && !p[a - 1]) {
          p[a - 1] = 1;
        } else {
          q.push_back(a);
        }
      }
      bool ff = true;
      sort(q.begin(), q.end());
      int pow = 0;
      for (int i = 0; i < n; i++) {
        if (!p[i]) {
          if (q[pow++] <= 2 * (i + 1)) {
            ff = false;
            break;
          }
        }
      }
      if (ff) {
        cout << q.size() << "\n";
      } else {
        cout << -1 << "\n";
      }
    }
    return 0;
  }
  