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.*;
public class Q1617C {

	public static void main(String[] args) {

		Scanner scn = new Scanner(System.in);
        int t = scn.nextInt();
        while(t-->0){
            int n = scn.nextInt();
            int[] arr = new int[n];
            for(int i=0;i<n;i++){
                int a = scn.nextInt();
                arr[i] = a;
            }
            System.out.println(solve(arr));
        }

	}
	public static int solve(int[] arr) {
		int n = arr.length;
		int[] ans = new int[n];
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>((a,b)->b-a);
        for(int i=0;i<n;i++) {
        	if(arr[i]<=n && ans[arr[i]-1]==0) 
        		ans[arr[i]-1] = arr[i];  
        	else
        		pq.add(arr[i]);
        }
        int an = pq.size();
        for(int i=n-1;i>=0;i--) {
        	if(ans[i]==0) {
        		int rem = pq.remove();
        		int pos = i+1;        		
        		if(rem%2==0) {
        			rem/=2;
        			rem--;
        		}else {
        			rem/=2;
        		}
        		if(rem<pos)
        			return -1;
        	}
        }
        return an;
	}
}
