#include <bits/stdc++.h>
using namespace std;
long long fun(long long n) {
  long long maxPrime = -1;
  while (n % 2 == 0) {
    maxPrime = 2;
    n >>= 1;
  }
  while (n % 3 == 0) {
    maxPrime = 3;
    n = n / 3;
  }
  for (int i = 5; i <= sqrt(n); i += 6) {
    while (n % i == 0) {
      maxPrime = i;
      n = n / i;
    }
    while (n % (i + 2) == 0) {
      maxPrime = i + 2;
      n = n / (i + 2);
    }
  }
  if (n > 4) maxPrime = n;
  return maxPrime;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v(n);
    vector<int> a;
    unordered_map<int, int> m;
    bool flag = false;
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      if (v[i] <= n && m.find(v[i]) == m.end())
        m[v[i]]++;
      else
        a.push_back(v[i]);
    }
    sort(a.begin(), a.end());
    int i = 1, j = 0;
    while (i <= n) {
      if (m.find(i) != m.end())
        i++;
      else {
        int d = (a[j] % 2 == 0) ? (a[j] / 2 - 1) : (a[j] / 2);
        if (i > d) {
          flag = true;
          break;
        } else {
          j++;
          i++;
          cnt++;
        }
      }
    }
    if (flag == true)
      cout << -1 << "\n";
    else
      cout << cnt << "\n";
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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>
using namespace std;
long long fun(long long n) {
  long long maxPrime = -1;
  while (n % 2 == 0) {
    maxPrime = 2;
    n >>= 1;
  }
  while (n % 3 == 0) {
    maxPrime = 3;
    n = n / 3;
  }
  for (int i = 5; i <= sqrt(n); i += 6) {
    while (n % i == 0) {
      maxPrime = i;
      n = n / i;
    }
    while (n % (i + 2) == 0) {
      maxPrime = i + 2;
      n = n / (i + 2);
    }
  }
  if (n > 4) maxPrime = n;
  return maxPrime;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v(n);
    vector<int> a;
    unordered_map<int, int> m;
    bool flag = false;
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      if (v[i] <= n && m.find(v[i]) == m.end())
        m[v[i]]++;
      else
        a.push_back(v[i]);
    }
    sort(a.begin(), a.end());
    int i = 1, j = 0;
    while (i <= n) {
      if (m.find(i) != m.end())
        i++;
      else {
        int d = (a[j] % 2 == 0) ? (a[j] / 2 - 1) : (a[j] / 2);
        if (i > d) {
          flag = true;
          break;
        } else {
          j++;
          i++;
          cnt++;
        }
      }
    }
    if (flag == true)
      cout << -1 << "\n";
    else
      cout << cnt << "\n";
  }
  return 0;
}

```