#include <bits/stdc++.h>
int main() {
  int tasks;
  std::cin >> tasks;
  for (int o = 0; o < tasks; ++o) {
    int n;
    std::cin >> n;
    int res = 0;
    std::vector<int> a(n);
    std::vector<int> b(n);
    std::vector<int> need_change;
    for (int i = 0; i < n; ++i) {
      std::cin >> a[i];
      if (a[i] <= n && b[a[i] - 1] == 0) {
        b[a[i] - 1] = 1;
      } else {
        need_change.push_back(a[i]);
      }
    }
    std::vector<int> change_to;
    for (int i = 0; i < n; ++i) {
      if (b[i] == 0) {
        change_to.push_back(i + 1);
      }
    }
    std::sort(need_change.begin(), need_change.end());
    bool can_we = 1;
    for (int i = 0; i < need_change.size(); ++i) {
      int max_get_from_op = need_change[i] / 2;
      if (need_change[i] % 2 == 1) {
        max_get_from_op += 1;
      }
      if (change_to[i] >= max_get_from_op) {
        can_we = 0;
      }
    }
    if (can_we) {
      std::cout << need_change.size() << std::endl;
    } else {
      std::cout << -1 << std::endl;
    }
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  int tasks;
  std::cin >> tasks;
  for (int o = 0; o < tasks; ++o) {
    int n;
    std::cin >> n;
    int res = 0;
    std::vector<int> a(n);
    std::vector<int> b(n);
    std::vector<int> need_change;
    for (int i = 0; i < n; ++i) {
      std::cin >> a[i];
      if (a[i] <= n && b[a[i] - 1] == 0) {
        b[a[i] - 1] = 1;
      } else {
        need_change.push_back(a[i]);
      }
    }
    std::vector<int> change_to;
    for (int i = 0; i < n; ++i) {
      if (b[i] == 0) {
        change_to.push_back(i + 1);
      }
    }
    std::sort(need_change.begin(), need_change.end());
    bool can_we = 1;
    for (int i = 0; i < need_change.size(); ++i) {
      int max_get_from_op = need_change[i] / 2;
      if (need_change[i] % 2 == 1) {
        max_get_from_op += 1;
      }
      if (change_to[i] >= max_get_from_op) {
        can_we = 0;
      }
    }
    if (can_we) {
      std::cout << need_change.size() << std::endl;
    } else {
      std::cout << -1 << std::endl;
    }
  }
}

```