#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n, match = 0;
    cin >> n;
    vector<long long int> a(n), v(n + 1);
    map<long long int, long long int> m;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      if (a[i] <= n && v[a[i]] == 0) match++, v[a[i]]++;
    }
    sort(a.begin(), a.end());
    for (int i = 1; i <= n; ++i) {
      if (binary_search(a.begin(), a.end(), i)) {
        m[upper_bound(a.begin(), a.end(), i) - a.begin() - 1] = 1;
      }
    }
    long long int ptr = n, i = n - 1, c = 0;
    while (ptr && i >= 0) {
      if (v[ptr])
        ptr--;
      else {
        while (m[i]) i--;
        if (a[i] <= 2 * ptr) {
          cout << -1 << '\n';
          c++;
          break;
        } else {
          v[ptr] = 1;
          ptr--;
          m[i] = 1;
        }
      }
    }
    if (c == 0) {
      cout << n - match << '\n';
    }
  }
}

### 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>
using namespace std;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n, match = 0;
    cin >> n;
    vector<long long int> a(n), v(n + 1);
    map<long long int, long long int> m;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      if (a[i] <= n && v[a[i]] == 0) match++, v[a[i]]++;
    }
    sort(a.begin(), a.end());
    for (int i = 1; i <= n; ++i) {
      if (binary_search(a.begin(), a.end(), i)) {
        m[upper_bound(a.begin(), a.end(), i) - a.begin() - 1] = 1;
      }
    }
    long long int ptr = n, i = n - 1, c = 0;
    while (ptr && i >= 0) {
      if (v[ptr])
        ptr--;
      else {
        while (m[i]) i--;
        if (a[i] <= 2 * ptr) {
          cout << -1 << '\n';
          c++;
          break;
        } else {
          v[ptr] = 1;
          ptr--;
          m[i] = 1;
        }
      }
    }
    if (c == 0) {
      cout << n - match << '\n';
    }
  }
}

```