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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.*;


 
public class Codeforces {
	static int mod= 1000000007;
	
	public static void main(String[] args) throws Exception {
		PrintWriter out=new PrintWriter(System.out);
		FastScanner fs=new FastScanner();
		int t=fs.nextInt();
		
		outer:while(t-->0) {
			int n=fs.nextInt();
			int arr[]=fs.readArray(n);
			sort(arr);
			int cur=n;
			int ans=0;
			for(int i=n-1;i>=0;i--) {
				int ele=arr[i];
				if(ele==cur) {
					cur--;
					continue;
				}
//				System.out.println(ele+" "+(ele-cur));
//				System.out.println(ele%(ele-cur)+" "+cur);
				if(ele-cur>1&&(ele%(ele-cur))==cur) {
					
					ans++;
					cur--;
				}
				else {
					out.println(-1);
					continue outer;
				}
				
				
			}
			out.println(ans);
		}
		
		out.close();
		
	}
	
	static long pow(long a,long b) {
		if(b<0) return 1;
		long res=1;
		while(b!=0) {
			if((b&1)!=0) {
				res*=a;
				res%=mod;
			}
			a*=a;
			a%=mod;
			b=b>>1;
		}
		return res;
	}
	static int gcd(int  a,int  b) {
		if(b==0) return a;
		return gcd(b,a%b);
	}
	static long nck(int n,int k) {
		if(k>n) return 0;
		long res=1;
		res*=fact(n);
		res%=mod;
		res*=modInv(fact(k));
		res%=mod;
		res*=modInv(fact(n-k)); 
		res%=mod;
		return res;
	}
	static long fact(long n) {
//		return fact[(int)n];
		long res=1;
		for(int i=2;i<=n;i++) {
			res*=i;
			res%=mod;
		}
		return res;
	}
	
	static long modInv(long n) {
		return pow(n,mod-2);
	}
	
	static void sort(int[] a) {
		//suffle
		int n=a.length;
		Random r=new Random();
		for (int i=0; i<a.length; i++) {
			int oi=r.nextInt(n);
			int temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		
		//then sort
		Arrays.sort(a);
	}
	
	// Use this to input code since it is faster than a Scanner
	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());
		}
		long[] lreadArray(int n) {
			long a[]=new long[n];
			for(int i=0;i<n;i++) a[i]=nextLong();
			return a;
		}
		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());
		}
	}
 
}