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 Main {
 
    public static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
 
    }
 
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        tc:while(t-->0) {
            int n = sc.nextInt();
            int [] arr = new int[n];
            boolean [] visited = new boolean[n+1];
            ArrayList<Integer> ans = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                arr[i] = sc.nextInt();
                if(arr[i]<=n && !visited[arr[i]]){
                    visited[arr[i]] = true;
                }else{
                    ans.add(arr[i]);
                }
            }
 
            Collections.sort(ans);
            int cnt = 0;
            int j = 0;
 
            for (int i = 1; i <= n; i++) {
 
                if(!visited[i]){
                    int temp = (ans.get(j)+1)/2;
                    j++;
                    if(temp<=i){
                        System.out.println(-1);
                        continue tc;
                    }
                    else cnt++;
 
                }
 
 
            }
            System.out.println(cnt);
 
            
        }
 
    }
 
}