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.*;
import java.io.*;
 
public class C {
	
	static FastScanner fs = new FastScanner();
	static PrintWriter pw = new PrintWriter(System.out);
	static StringBuilder sb = new StringBuilder("");
	
	public static void main(String[] args) {
		int t = fs.nextInt();
		for (int tt = 0; tt < t; tt++) {
			int n = fs.nextInt();
			int[] a = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = fs.nextInt();
				a[i]--;
			}
			boolean[] used = new boolean[n];
			ArrayList<Integer> b = new ArrayList<>();
			for (int i = 0; i < n; i++) {
				if (a[i] >= n) b.add(a[i]);
				else if (!used[a[i]]) used[a[i]] = true;
				else b.add(a[i]);
			}
			int m = b.size();
			Collections.sort(b);
			int cnt = 0;
			boolean ok = true;
			for (int i = 0; i < n; i++) {
				if (!used[i]) {
					if (i >= b.get(cnt) / 2) {
						ok = false;
					}
				cnt++;
				}
			}
			if(ok) sb.append(m + "\n");
			else sb.append(-1 + "\n");
		}
		pw.print(sb.toString());
		pw.close();
	}
	
	static void sort(long[] a) {
		ArrayList<Long> al = new ArrayList<>();
		for (long i : a)
			al.add(i);
		Collections.sort(al);
		for (int i = 0; i < a.length; i++)
			a[i] = al.get(i);
	}
 
	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) {}return st.nextToken();}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());}
		double nextDouble() {return Double.parseDouble(next());}
		int[] readArray(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) {a[i] = nextInt();} return a;}
	}
}