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;
  const int inf = 0x3f3f3f3f;
  const long long INF = 0x3f3f3f3f3f3f3f3f;
  inline long long read() {
    long long first = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      first = first * 10 + ch - '0';
      ch = getchar();
    }
    return first * f;
  }
  inline void ou(long long first) {
    char ch[40];
    int len = 0;
    if (first < 0) {
      putchar('-');
      first = ~first + 1;
    }
    do {
      ch[len++] = first % 10 + '0';
      first /= 10;
    } while (first > 0);
    for (int i = len - 1; i >= 0; i--) putchar(ch[i]);
    return;
  }
  inline int addmod(int u_, int v_, int MOD) {
    u_ += v_;
    return u_ >= MOD ? u_ - MOD : u_;
  }
  inline int submod(int u_, int v_, int MOD) {
    u_ -= v_;
    return u_ < 0 ? u_ + MOD : u_;
  }
  const double eps = 1e-8;
  const int maxn = 1e5 + 100;
  const int mod = 1e9 + 7;
  const int N = 1010;
  int a[maxn], st[maxn];
  void solve() {
    std::vector<int> v;
    int n = read();
    for (int i = 1; i <= n; ++i) st[i] = 0;
    for (int i = 1; i <= n; ++i) {
      a[i] = read();
      if (a[i] <= n and !st[a[i]])
        st[a[i]] = 1;
      else
        v.push_back(a[i]);
    }
    vector<int> b;
    for (int i = 1; i <= n; ++i)
      if (!st[i]) b.push_back(i);
    sort((v).begin(), (v).end());
    int si = v.size();
    for (int i = 0; i <= si - 1; ++i) {
      int dod = 0;
      if (v[i] & 1)
        dod = v[i] / 2;
      else
        dod = v[i] / 2 - 1;
      if (dod < b[i]) {
        ou(-1), puts("");
        return;
      }
    }
    ou(si), puts("");
  }
  signed main() {
    int _ = 1;
    _ = read();
    for (int cas = 1; cas <= _; ++cas) {
      solve();
    }
    return 0;
  }
  