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;
template <class T1, class T2>
void maximize(T1 &a, T2 b) {
  if (b > a) a = b;
}
template <class T1, class T2>
void minimize(T1 &a, T2 b) {
  if (b < a) a = b;
}
template <class T>
void read(T &number) {
  bool negative = false;
  register int c;
  number = 0;
  c = getchar();
  while (c != '-' && !isalnum(c)) c = getchar();
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
  if (negative) number *= -1;
}
template <class T, class... Ts>
void read(T &a, Ts &...args) {
  read(a);
  read(args...);
}
int n;
int a[1000001];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int _ = (1); _ <= (t); _++) {
    cin >> n;
    for (int i = (1); i <= (n); i++) cin >> a[i];
    sort(a + 1, a + 1 + n);
    int ans = 0;
    for (int i = (1); i <= (n); i++) {
      if (i == a[i]) continue;
      ans++;
      int val = a[i] / 2 - (a[i] % 2 == 0);
      if (val < i) {
        ans = -1;
        break;
      }
    }
    cout << ans << '\n';
  }
}
