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

/*
MMMMMMMM               MMMMMMMM        KKKKKKKKK    KKKKKKK
M:::::::M             M:::::::M        K:::::::K    K:::::K
M::::::::M           M::::::::M        K:::::::K    K:::::K
M:::::::::M         M:::::::::M        K:::::::K   K::::::K
M::::::::::M       M::::::::::M        KK::::::K  K:::::KKK
M:::::::::::M     M:::::::::::M          K:::::K K:::::K
M:::::::M::::M   M::::M:::::::M          K::::::K:::::K
M::::::M M::::M M::::M M::::::M          K:::::::::::K
M::::::M  M::::M::::M  M::::::M          K:::::::::::K
M::::::M   M:::::::M   M::::::M          K::::::K:::::K
M::::::M    M:::::M    M::::::M          K:::::K K:::::K
M::::::M     MMMMM     M::::::M        KK::::::K  K:::::KKK
M::::::M               M::::::M        K:::::::K   K::::::K
M::::::M               M::::::M ...... K:::::::K    K:::::K
M::::::M               M::::::M .::::. K:::::::K    K:::::K
MMMMMMMM               MMMMMMMM ...... KKKKKKKKK    KKKKKKK
*/
#include <bits\stdc++.h>

using namespace std;

#define ll long long
#define dd double

int main()
{
	//ios_base::sync_with_stdio(0);
	//cin.tie(0);
	//cout.tie(0);
//-------------------------//

	int t;cin >> t;
	while (t--)
	{
		ll n;cin >> n;
		vector < ll > v(n + 1), ch(n + 1);
		for (int i = 1;i <= n;i++)
		{
			cin >> v[i];
		}
		sort(v.begin(), v.end());
		ll ans = n;
		for (int i = 1;i <= n;i++)
		{
			if (v[i] == i)
			{
				ans--;
			}
			else
				break;
		}
		for (int i = n - ans + 1;i <= n;i++)
		{
			ll x = min(n, v[i] / 2);
			ch[x]++;
		}
		ll cnt = 0;
		bool ok = 1;
		for (int i = n;i > (n - ans);i--)
		{
			cnt += ch[i];
			if (cnt == 0)
				ok = 0;
		}
		if (ok)
		{
			cout << ans << endl;
		}
		else
			cout << "-1" << endl;
	}
	return 0;
}