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 void main(String args[])
  {
    int i,count;
    Scanner sc=new Scanner(System.in);
    int t=sc.nextInt();
    while(t-- >0)
    {
      Set<Integer> s1=new HashSet<Integer>();
      count=0;
      int n=sc.nextInt();
      Integer []arr=new Integer[n];
      for(i=0;i<n;i++)
      {
        arr[i]=sc.nextInt();
        if(arr[i]<=n && !s1.contains(arr[i]))
        {
          s1.add(arr[i]);
          arr[i]=-1;
        }
      }
      i=0;
      Arrays.sort(arr,Collections.reverseOrder());
      while(n>=1)
      {
        if(s1.contains(n))
        {
          n--;
          continue;
        }
        else{
          if(arr[i]==-1)
          {
            count=-1;
            break;
          }
          else if(arr[i]-n>1 && arr[i]%(arr[i]-n)!=0)
          {
            i++;
            count++;
            n--;
          }
          else{
            count=-1;
            break;
          }
        }
      }
      System.out.println(count);
    }
  }
}