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;
long long int mod = 1e9 + 7;
long long int mxm = 1e18;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n], i;
    set<long long int> x;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      x.insert(a[i]);
    }
    vector<long long int> v(n, 0);
    for (auto xx : x) {
      if (xx <= n) v[xx - 1] = xx;
    }
    long long int ans = 0;
    vector<long long int> w;
    for (i = 0; i < n; i++) {
      if (a[i] > n) {
        w.push_back(a[i]);
      }
    }
    sort((w).begin(), (w).end());
    long long int j = 0, m = w.size();
    for (i = 0; i < n && j < m; i++) {
      if (v[i] == 0) {
        if (w[j] > 2 * (i + 1)) {
          ans++;
        } else {
          ans = -1;
          break;
        }
        j++;
      }
    }
    sort(a, a + n);
    long long int ans1 = 0;
    for (i = 0; i < n; i++) {
      if (a[i] != i + 1) {
        if (a[i] > 2 * (i + 1)) {
          ans1++;
        } else {
          ans1 = -1;
          break;
        }
      }
    }
    if (ans == -1 && ans1 == -1)
      cout << (-1) << endl;
    else if (ans == -1)
      cout << (ans1) << endl;
    else if (ans1 == -1)
      cout << (ans) << endl;
    else
      cout << (min(ans, ans1)) << endl;
  }
  return 0;
}
