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 Codeforces {


	// CHECK CONSTRAINTS ALWAYS EDGE CASE MISSED 
	
static int mod = 1000000007;	

static class Pair{
	long a;
	long b;
	Pair(long a,  long b){
		this.a = a;
		this.b = b;
	}
}

public static void main(String[] args) {

Scanner sc=  new Scanner(System.in);

  while(sc.hasNext()) {

int t  =  sc.nextInt();
 while(t-->0) {
	
int n  = sc.nextInt();
ArrayList<Integer> al = new ArrayList<>();
ArrayList<Integer> bl = new ArrayList<>();
 int arr[] = new int[n+1];
for(int i = 0 ; i <n;i++) {
	int a = sc.nextInt();
	if(a>n) {
		al.add(a);
	}
	else {
		arr[a]++;
	}
}

for(int i =1 ; i<=n ; i++) {
	if(arr[i] == 0) {
		bl.add(i);
	}
	if(arr[i]>1) {
		while(arr[i] -- >1) {
			al.add(i);
			}
		
	}
}
Collections.sort(al);
Collections.sort(bl);

boolean flag = true;
long cnt = 0;
for(int i = 0; i <al.size();  i++) {
	
	long mid = (al.get(i) + 1)/2;
	if(bl.get(i)>=mid)
	{
	    flag = false;
	    break;
	}
	else {
		cnt++;
	}
	
}


if(flag) {
	System.out.println(cnt);
}
else {
	System.out.println(-1);
}














  
  
	
		 
	 
	
     }

 
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}
 

 
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