#include <bits/stdc++.h>
using namespace std;
long long int Mod = 1e9 + 7;
vector<bool> status(1e6 + 1, 1);
vector<long long int> primes;
void test_case() {
  int n, x;
  cin >> n;
  int hash[n + 1];
  memset(hash, 0, sizeof hash);
  vector<int> a;
  for (int i = 0; i < n; i++) {
    cin >> x;
    if (x <= n)
      hash[x]++;
    else
      a.push_back(x);
  }
  int ans[n + 1];
  memset(ans, 0, sizeof ans);
  for (int i = 1; i < n + 1; i++) {
    if (hash[i] > 0) {
      ans[i] = i;
      hash[i]--;
    }
    while (hash[i] > 0) {
      a.push_back(i);
      hash[i]--;
    }
  }
  int j = 0, cnt = a.size();
  sort(a.begin(), a.end());
  for (int i = 1; i < n + 1; i++) {
    if (ans[i] == 0) {
      if (a[j] >= 2 * i + 1) {
        ans[i] = i;
        j++;
      } else {
        cout << -1 << '\n';
        return;
      }
    }
  }
  cout << cnt << '\n';
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  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;
long long int Mod = 1e9 + 7;
vector<bool> status(1e6 + 1, 1);
vector<long long int> primes;
void test_case() {
  int n, x;
  cin >> n;
  int hash[n + 1];
  memset(hash, 0, sizeof hash);
  vector<int> a;
  for (int i = 0; i < n; i++) {
    cin >> x;
    if (x <= n)
      hash[x]++;
    else
      a.push_back(x);
  }
  int ans[n + 1];
  memset(ans, 0, sizeof ans);
  for (int i = 1; i < n + 1; i++) {
    if (hash[i] > 0) {
      ans[i] = i;
      hash[i]--;
    }
    while (hash[i] > 0) {
      a.push_back(i);
      hash[i]--;
    }
  }
  int j = 0, cnt = a.size();
  sort(a.begin(), a.end());
  for (int i = 1; i < n + 1; i++) {
    if (ans[i] == 0) {
      if (a[j] >= 2 * i + 1) {
        ans[i] = i;
        j++;
      } else {
        cout << -1 << '\n';
        return;
      }
    }
  }
  cout << cnt << '\n';
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    test_case();
  }
}

```