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;

import static java.lang.Math.max;

public class Main {

    public static int solution(int n, ArrayList<Integer> t) {
        int ans = 0;
        Collections.sort(t);
//        for (int i = 0; i < n; i++) {
//            if (t.get(i) == i + 1)
//                continue;
//
//            int a = t.get(i), b = i + 1, c = a - b;
//            if (c <= b)
//                return -1;
//            ans++;
//        }

        int k = 0;
        int j = 0;
        ArrayList<Integer> used = new ArrayList();
        int next_unused = 0;
        for (int i = 0; i < n; i++)
            used.add(0);
        for (int i = 0; i < n; i++) {
            int b = i + 1;
            while(k < n && t.get(k) < b)
                k++;
            while(j < n && (t.get(j) < 2 * b + 1 || used.get(j) == 1))
                j++;

            next_unused = max(next_unused, k);
            while(next_unused < n && used.get(next_unused) == 1) {
                next_unused++;
                continue;
            }
            while(next_unused >= n && k < n && t.get(k) <= b)
                k++;

            //System.err.println("" + b + " " + t.get(k) + " " + t.get(j));
            if (k < n && t.get(k) == b) {
                if (used.get(k) == 1 && next_unused < n) {
                    used.set(next_unused, 1);
                    next_unused++;
                    continue;
                }

                used.set(k, 1);
                k++;
                continue;
            }
            if (j < n && t.get(j) >= 2 * b + 1) {
                used.set(j, 1);
                j++;
                ans++;
                continue;
            }
            return -1;
        }

        return ans;
    }

    public static void main(String[] args) {
	// write your code here
        Scanner in = new Scanner(System.in);
        int Q = in.nextInt();
        in.nextLine();
        for (int q = 0; q < Q; q++) {
            int n = in.nextInt();
            ArrayList<Integer> t = new ArrayList();
            for (int i = 0 ; i < n; i++) {
                int a = in.nextInt();
                t.add(a);
            }
            System.out.println(solution(n, t));
        }
    }
}
