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;

#define ll long long
#define pb push_back
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vi vector<int> 
#define vll(x) vector<ll> x
#define rep(x,s,e) for(ll x = s ; x < e ; ++x)
#define nl '\n';
 

int main(){
	int t;
	cin>>t;
	
	while(t--){
		int n,x;
		cin>>n;
		
		vi arr;
		bool mark[n+2] = {0};
		
		for (int i = 0; i < n; i++)
		{
			cin>>x;
			arr.pb(x);
			
			if(x <= n)
				mark[x] = 1;
		}
		
		sort(arr.begin() , arr.end());
		
		int s = 1 , j = 0 , res = 0;
		
		
		while(s <= n && j < n){
			if( (arr[j] > n || mark[arr[j]] == 0) && arr[j] <= 2*s ){
				res = -1;
				break;
			}
			else{
				if(arr[j] <= n && mark[arr[j]]){
					
					mark[arr[j]] = 0;
					
					if(arr[j] == s)
						++s;
					
					++j;
					
				}
				
					
					
				else {
					++j;
					++s;
					++res;
				}	
				
			}
		}
		cout<<res<<nl;
		
	}

return 0;
}
