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

//Paprika and Permutation
import java.io.*;
import java.util.*;

public class C1617{
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      int t = Integer.parseInt(f.readLine());
      
      for(int q = 1; q <= t; q++){

         int n = Integer.parseInt(f.readLine());
      
         StringTokenizer st = new StringTokenizer(f.readLine());
         Integer[] array = new Integer[n];
         
         HashSet<Integer> hset = new HashSet<Integer>();
         ArrayList<Integer> notused = new ArrayList<Integer>();
         for(int k = 0; k < n; k++){
            array[k] = Integer.parseInt(st.nextToken());
            
            if(array[k] > n || hset.contains(array[k])){
               notused.add(array[k]);
            } else {
               hset.add(array[k]);
            }
         }
      
         Collections.sort(notused);
         int nui = 0;
         
         boolean fail = false;
         int answer = 0;
         for(int k = 1; k <= n; k++){
            if(hset.contains(k)) continue;
            
            if((notused.get(nui)-1)/2<k){
               fail = true;
               break;
            } else {
               answer++;
            }
            
            nui++;
         }
         
         if(fail){
            out.println(-1);
         } else {
            out.println(answer);
         }
      }
      
      
      
      
      out.close();
   }
   
      
}