#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
const int SZ = 3e5 + 100;
void initialize() {}
void test_case() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (auto &x : arr) {
    cin >> x;
    x -= 1;
  }
  vector<int> per(n + 1), e;
  for (int i = 0; i < n; i++) {
    if (arr[i] < n and per[arr[i]] == 0) {
      per[arr[i]] = 1;
    } else {
      e.push_back(arr[i]);
    }
  }
  vector<int> req;
  for (int i = 0; i < n; i++)
    if (per[i] == 0) req.push_back(i);
  assert(e.size() == req.size());
  sort(e.begin(), e.end());
  for (int i = req.size() - 1; i >= 0; i--) {
    int a = e.back() + 1;
    int b = req[i] + 1;
    int x = a - b;
    if (e.back() < req[i] or x == 0 or a % x != b) {
      cout << "-1\n";
      return;
    }
    e.pop_back();
  }
  cout << req.size() << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  srand(time(0));
  initialize();
  int t;
  cin >> t;
  while (t--) {
    test_case();
  }
}

### Prompt

Create a solution in cpp for 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;
const long long MOD = 998244353;
const int SZ = 3e5 + 100;
void initialize() {}
void test_case() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (auto &x : arr) {
    cin >> x;
    x -= 1;
  }
  vector<int> per(n + 1), e;
  for (int i = 0; i < n; i++) {
    if (arr[i] < n and per[arr[i]] == 0) {
      per[arr[i]] = 1;
    } else {
      e.push_back(arr[i]);
    }
  }
  vector<int> req;
  for (int i = 0; i < n; i++)
    if (per[i] == 0) req.push_back(i);
  assert(e.size() == req.size());
  sort(e.begin(), e.end());
  for (int i = req.size() - 1; i >= 0; i--) {
    int a = e.back() + 1;
    int b = req[i] + 1;
    int x = a - b;
    if (e.back() < req[i] or x == 0 or a % x != b) {
      cout << "-1\n";
      return;
    }
    e.pop_back();
  }
  cout << req.size() << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  srand(time(0));
  initialize();
  int t;
  cin >> t;
  while (t--) {
    test_case();
  }
}

```