#include <bits/stdc++.h>
using namespace std;
long long mod(long long x, long long MOD) {
  if (x > MOD) x -= MOD;
  return x;
}
long long _ceil(long long x, long long y) { return x / y + (x % y != 0); }
void wrt() {
  cout << "\n";
  return;
}
template <typename T1, typename... T2>
void wrt(T1 x, T2... args) {
  ((cout << x << ' '), wrt(args...));
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto& i : v) os << i << " ";
  return os;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& v) {
  for (auto& e : v) in >> e;
  return in;
}
const long long MOD = 1e9 + 7;
const long long N = 1e6 + 5, inf = 1e18 + 2;
long long n, m, x, u, v, k, p, q, i, j;
string s, t;
long long ans;
long long a[N];
void solve() {
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  multiset<long long> st((a).begin(), (a).end());
  x = 0;
  vector<long long> vis(n + 1, 0);
  for (long long i = 1; i <= (n); ++i) {
    if (st.find(i) != st.end()) {
      st.erase(st.find(i));
      vis[i] = 1;
    }
  }
  for (long long i = 1; i <= (n); ++i) {
    if (!vis[i]) {
      if (*st.begin() - i > i) {
        vis[i] = 1;
        st.erase(st.begin());
      }
      x++;
    }
  }
  for (long long i = 1; i <= (n); ++i)
    if (!vis[i]) return wrt(-1);
  wrt(x);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
long long mod(long long x, long long MOD) {
  if (x > MOD) x -= MOD;
  return x;
}
long long _ceil(long long x, long long y) { return x / y + (x % y != 0); }
void wrt() {
  cout << "\n";
  return;
}
template <typename T1, typename... T2>
void wrt(T1 x, T2... args) {
  ((cout << x << ' '), wrt(args...));
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto& i : v) os << i << " ";
  return os;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& v) {
  for (auto& e : v) in >> e;
  return in;
}
const long long MOD = 1e9 + 7;
const long long N = 1e6 + 5, inf = 1e18 + 2;
long long n, m, x, u, v, k, p, q, i, j;
string s, t;
long long ans;
long long a[N];
void solve() {
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  multiset<long long> st((a).begin(), (a).end());
  x = 0;
  vector<long long> vis(n + 1, 0);
  for (long long i = 1; i <= (n); ++i) {
    if (st.find(i) != st.end()) {
      st.erase(st.find(i));
      vis[i] = 1;
    }
  }
  for (long long i = 1; i <= (n); ++i) {
    if (!vis[i]) {
      if (*st.begin() - i > i) {
        vis[i] = 1;
        st.erase(st.begin());
      }
      x++;
    }
  }
  for (long long i = 1; i <= (n); ++i)
    if (!vis[i]) return wrt(-1);
  wrt(x);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```