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.Scanner;

/**
 *
 * @author Acer
 */
public class PaprikaAndPermutation_C {
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while(T-- > 0){
            int n = sc.nextInt();
            ArrayList<Integer> arr = new ArrayList<Integer>();
            ArrayList<Integer> A = new ArrayList<Integer>();
            ArrayList<Integer> B = new ArrayList<Integer>();
            for (int i = 1; i <= n; i++) {
                int a = sc.nextInt();
                if(!arr.contains(a) && a <= n){
                    arr.add(a);
                }
                else{
                    A.add(a);
                    B.add(i);
                }
            }
            
            if(A.isEmpty()){
                System.out.println(0);
            }
            else{
                Collections.sort(A);
                boolean flag = true;
                for (int i = 0; i < A.size(); i++) {
                    if((A.get(i)+1)/2 <= B.get(i)){
                       flag = false;
                    }
                }
                if(flag){
                    System.out.println(A.size());
                }
                else{
                    System.out.println(-1);
                }
            }
        }
    }
    
}