#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int test_case, n;
  cin >> test_case;
  while (test_case--) {
    cin >> n;
    long long int ar[n], i, c = 0, b, j = 0;
    vector<long long int> w;
    vector<pair<long long int, long long int>> v;
    map<long long int, long long int> m, p;
    for (i = 0; i < n; i++) {
      cin >> ar[i];
      m[ar[i]]++;
    }
    sort(ar, ar + n);
    for (i = 1; i <= n; i++) {
      p[i] = 1;
      if (m[i] == 0) {
        w.push_back(i);
      }
    }
    for (i = 0; i < n; i++) {
      if (p[ar[i]] == 0) {
        v.push_back({ar[i], w[j]});
        j++;
      } else {
        p[ar[i]] = 0;
      }
    }
    for (i = 0; i < v.size(); i++) {
      b = v[i].first - v[i].second;
      if (b <= v[i].second) {
        c = 1;
        break;
      }
    }
    if (c) {
      cout << "-1" << endl;
    } else {
      cout << v.size() << endl;
    }
  }
}

### 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int test_case, n;
  cin >> test_case;
  while (test_case--) {
    cin >> n;
    long long int ar[n], i, c = 0, b, j = 0;
    vector<long long int> w;
    vector<pair<long long int, long long int>> v;
    map<long long int, long long int> m, p;
    for (i = 0; i < n; i++) {
      cin >> ar[i];
      m[ar[i]]++;
    }
    sort(ar, ar + n);
    for (i = 1; i <= n; i++) {
      p[i] = 1;
      if (m[i] == 0) {
        w.push_back(i);
      }
    }
    for (i = 0; i < n; i++) {
      if (p[ar[i]] == 0) {
        v.push_back({ar[i], w[j]});
        j++;
      } else {
        p[ar[i]] = 0;
      }
    }
    for (i = 0; i < v.size(); i++) {
      b = v[i].first - v[i].second;
      if (b <= v[i].second) {
        c = 1;
        break;
      }
    }
    if (c) {
      cout << "-1" << endl;
    } else {
      cout << v.size() << endl;
    }
  }
}

```