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.
8
2
2 3
3
3 4 5
1
1
1
2
1
3
1
4
1
12071207
5
5 5 5 9 14

1
2
0
-1
1
1
1
4

#include <bits/stdc++.h>
using namespace std;
bool spc;
template <typename T, typename U>
istream& operator>>(istream& i, pair<T, U>& p) {
  i >> p.first >> p.second;
  return i;
}
template <typename T, typename U>
ostream& operator<<(ostream& o, pair<T, U>& p) {
  o << p.first << ' ' << p.second;
  return o;
}
template <typename T>
istream& operator>>(istream& i, vector<T>& v) {
  for (auto& el : v) i >> el;
  return i;
}
template <typename T>
ostream& operator<<(ostream& o, vector<T>& v) {
  for (auto& el : v) {
    spc = 0;
    cout << el;
    if (!spc) cout << ' ';
  }
  spc = 1;
  cout << '\n';
  return o;
}
void solve();
void solvet() {
  int t;
  cin >> t;
  for (long long __ = (0); __ < (t); __++) solve();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  ;
  solvet();
  return 0;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  sort((a).begin(), (a).end());
  long long ans = 0;
  for (long long i = (0); i < (n); i++) {
    if (a[i] != i + 1) ans++;
    if (a[i] < i + 1 || a[i] != i + 1 && (a[i] - (i + 1) <= i + 1 ||
                                          a[i] % (a[i] - i - 1) != i + 1)) {
      cout << -1 << endl;
      return;
    }
  }
  cout << ans << endl;
}
