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.HashMap;
import java.util.Arrays;
import java.util.Scanner;
public class C1617{
    
 public static int[] sort(int[] arr,int low,int high){
        int[] sortedArray;
        if (low==high) {
            sortedArray=new int[]{arr[low]};
            return sortedArray;
        }
        int mid=(low+high)/2;
        int[] arr1=sort(arr,low,mid);
        int[] arr2=sort(arr,mid+1,high);

        sortedArray=merge(arr1,low,mid,arr2,mid+1,high);
        return sortedArray;

    }

    public static int[] merge(int[] arr1,int low1,int high1,int[] arr2,int low2,int high2){

        int size1=high1-low1+1;
        int size2=high2-low2+1;
        
        int[] arr=new int[size1+size2];
        int pointer=0;
        low1=0;low2=0;
        while (low1<size1 && low2<size2){
            if (arr1[low1] <= arr2[low2]) {
                arr[pointer++]=arr1[low1++];
            }
            else{
                arr[pointer++]=arr2[low2++];
            }

        }

        while(low1<size1) arr[pointer++]=arr1[low1++];
        while(low2<size2) arr[pointer++]=arr2[low2++];
        return arr;
    }
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-- >0){
                int n=sc.nextInt();
                int arr[]=new int[n];
                for(int i=0;i<n;i++) arr[i]=sc.nextInt();
                arr=sort(arr,0,n-1);
                int next=1;
                int ans=0;
                int f[]=new int[n+1];
                for(int i=0;i<n;i++){
                    if (arr[i]==next) {
                        f[next]=1;
                        while(next <=n && f[next]==1) next++;
                    }
                    else{
                        if (arr[i] <= n && f[arr[i]]==0) {
                            f[arr[i]]=1;
                        }
                        else{
                            if (next < (arr[i]+1)/2) {
                                ans++;
                                f[next]=1;
                                while(next <=n && f[next]==1) next++;
                            }
                            else {
                                ans=-1;
                                break;
                            }
                        }
                    }
                }        
                System.out.println(ans);
        }
        sc.close();
    }   
}
