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;
typedef long long int ll;
#define f(a, b, c, d) for(ll a = b; a < c; a += d)
#define pb push_back
#define umap unordered_map
#define uset unordered_set
#define pq priority_queue

void solve()
{
	ll n;
	cin >> n;
	ll a[n + 1] = {};
	vector<ll> v;
	f(i, 0, n, 1)
	{
		ll num;
		cin >> num;
		if (num <= n)
		{
			if (a[num])
				v.pb(num);
			else
				a[num] = 1;
		}
		else
			v.pb(num);
	}
	if (v.empty())
	{
		cout << "0\n";
		return;
	}
	ll ans = 0;
	sort(v.begin(), v.end());
	for (ll i = n; i >= 1; --i)
	{
		if (a[i])
			continue;
		if (v.back() - i > v.back() / 2)
			ans++;
		else
		{
			cout << "-1\n";
			return;
		}
	}
	cout << ans << "\n";
}

signed main()
{

#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	ll t = 1;
	cin >> t;
	while (t--)
		solve();
}