#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const double pi = acosl(-1);
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < (n); i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  set<long long> s;
  for (long long i = 1; i < (n + 1); i++) s.insert(i);
  vector<long long> vis(n);
  for (long long i = 0; i < (n); i++) {
    long long x = a[i];
    if (x <= n && s.find(x) != s.end()) {
      s.erase(x);
      vis[i] = 1;
    }
  }
  long long count = 0;
  for (long long i = 0; i < (n); i++) {
    if (vis[i]) continue;
    long long x = a[i] / 2;
    if (a[i] % 2 == 0) x--;
    if (*s.begin() <= x) {
      count++;
      s.erase(s.begin());
    }
  }
  if (s.size() == 0) {
    cout << count << '\n';
  } else {
    cout << -1 << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tt = 1;
  cin >> tt;
  while (tt--) {
    solve();
  }
}

### Prompt

Please formulate a cpp solution to 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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const double pi = acosl(-1);
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < (n); i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  set<long long> s;
  for (long long i = 1; i < (n + 1); i++) s.insert(i);
  vector<long long> vis(n);
  for (long long i = 0; i < (n); i++) {
    long long x = a[i];
    if (x <= n && s.find(x) != s.end()) {
      s.erase(x);
      vis[i] = 1;
    }
  }
  long long count = 0;
  for (long long i = 0; i < (n); i++) {
    if (vis[i]) continue;
    long long x = a[i] / 2;
    if (a[i] % 2 == 0) x--;
    if (*s.begin() <= x) {
      count++;
      s.erase(s.begin());
    }
  }
  if (s.size() == 0) {
    cout << count << '\n';
  } else {
    cout << -1 << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tt = 1;
  cin >> tt;
  while (tt--) {
    solve();
  }
}

```