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

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class PaprikaandPermutation {
	public static void solve(int[] arr,int n) {
		HashSet<Integer> set=new HashSet<Integer>();
		ArrayList<Integer> al=new ArrayList<Integer>();
		for (int i = 0; i < arr.length; i++) {
			if(arr[i]<=n) {
				if(!set.contains(arr[i])) {
					set.add(arr[i]);
				}
				else {
					al.add(arr[i]);
				}
			}
			else{
				al.add(arr[i]);
			}
		}
		Collections.sort(al,Collections.reverseOrder());
		int count=al.size();
		for (int i = 1; i <=n; i++) {
			if(!set.contains(i)&&al.size() > 0) {
				int x=al.remove(al.size()-1);
				
//				System.out.println(x%(x-i)+" "+i+" "+x);
				if(x%(x-i)==i) {
					
					continue;
				}
				else {
					System.out.println(-1);
					return;
				}
			}
		}
		System.out.println(count);
	}
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in=new Scanner(System.in);
		int t=in.nextInt();
		for (int i = 0; i < t; i++) {
			int n=in.nextInt();
			int[] arr=new int[n];
			for (int j = 0; j < n; j++) {
				arr[j] = in.nextInt();
			}
			solve(arr, n);
		}
	}

}
