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.io.*;
import java.util.*;

public class Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int t=Integer.parseInt(bu.readLine());
        while(t-->0)
        {
            int n=Integer.parseInt(bu.readLine());
            int a[]=new int[n],i;
            String s[]=bu.readLine().split(" ");
            HashMap<Integer,Integer> hm=new HashMap<>();
            for(i=0;i<n;i++)
            {
                a[i]=Integer.parseInt(s[i]);
                hm.put(a[i],hm.getOrDefault(a[i],0)+1);
            }

            for(i=0;i<n;i++)
            if(hm.get(i+1)!=null)   //only 1 reduction per number will happen
            {
                a[i]=i+1;
                hm.put(a[i],hm.get(a[i])-1);
            }
            else a[i]=-1;

            PriorityQueue<Integer> pq=new PriorityQueue<>();
            for(int x:hm.keySet())
            {
                int v=hm.get(x);
                while(v-->0) pq.add(x);
            }

            int ans=0;
            for(i=0;i<n;i++)
            if(a[i]==-1)
            {
                a[i]=pq.poll();
                int mod=a[i]-(i+1);
                if(mod<=0 || a[i]%mod!=(i+1)) {ans=-1; break;}
                ans++;
            }
            sb.append(ans+"\n");
        }
        System.out.print(sb);
    }
}