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 C {
	
	public static void main(String[] args)throws IOException {
		
		FastScanner scan = new FastScanner();
		PrintWriter output = new PrintWriter(System.out);
		int t = scan.nextInt();
		for(int tt = 0;tt<t;tt++) {
			int n = scan.nextInt();
			int arr[] = scan.readArray(n);
			HashSet<Integer> seen = new HashSet<>();
			for(int i = 0;i<n;i++) {
				if(arr[i]<=n && !seen.contains(arr[i])) {
					seen.add(arr[i]);
					arr[i] = 0;
				}
			}
			sort(arr);
			int notSeen = 1;
			boolean possible = true;
			for(int i = 0;i<n;i++) {
				if(arr[i]==0) continue;
				if(seen.contains(notSeen)) {
					notSeen++;
					i--;
					continue;
				}
				if((arr[i]-1)/2 < notSeen) {
					possible = false;
					break;
				}
				notSeen++;
			}
			output.println(possible ? n-seen.size() : -1);
			
		}
		output.flush();
	}
	
	public static int[] sort(int arr[]) {
		List<Integer> list = new ArrayList<>();
		for(int i:arr) list.add(i);
		Collections.sort(list);
		for(int i = 0;i<list.size();i++) arr[i] = list.get(i);
		return arr;
	}
	
	public static long gcd(long a, long b) {
		if(a == 0) return b;
		return gcd(b%a, a);
	}
	
	public static void printArray(int arr[]) {
		for(int i:arr) System.out.print(i+" ");
		System.out.println();
	}
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}

}
