#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int arr[n];
    map<int, int> m;
    vector<int> extra;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      if (arr[i] <= n) {
        if (m[arr[i]]) {
          extra.push_back(arr[i]);
        } else {
          m[arr[i]]++;
        }
      } else {
        extra.push_back(arr[i]);
      }
    }
    vector<int> remain;
    for (int i = 1; i <= n; i++) {
      if (!m[i]) {
        remain.push_back(i);
      }
    }
    sort(remain.begin(), remain.end());
    sort(extra.begin(), extra.end());
    int flag = 0;
    for (int i = 0; i < remain.size(); i++) {
      if (remain[i] >= (float)extra[i] / 2) {
        flag = 1;
      }
    }
    if (flag) {
      cout << "-1\n";
    } else {
      cout << remain.size() << endl;
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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>
using namespace std;
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int arr[n];
    map<int, int> m;
    vector<int> extra;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      if (arr[i] <= n) {
        if (m[arr[i]]) {
          extra.push_back(arr[i]);
        } else {
          m[arr[i]]++;
        }
      } else {
        extra.push_back(arr[i]);
      }
    }
    vector<int> remain;
    for (int i = 1; i <= n; i++) {
      if (!m[i]) {
        remain.push_back(i);
      }
    }
    sort(remain.begin(), remain.end());
    sort(extra.begin(), extra.end());
    int flag = 0;
    for (int i = 0; i < remain.size(); i++) {
      if (remain[i] >= (float)extra[i] / 2) {
        flag = 1;
      }
    }
    if (flag) {
      cout << "-1\n";
    } else {
      cout << remain.size() << endl;
    }
  }
  return 0;
}

```