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

// Problem: C. Paprika and Permutation
// Contest: Codeforces - Codeforces Round #761 (Div. 2)
// URL: https://codeforces.com/contest/1617/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms

import java.io.*;
import java.util.*;

import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;



public class X1631 {
    long mod = 1000000007;
    private void preparation() {

    }


    private void clear() {

    }

    private void solve() throws Exception {
        rf();
        int n = ri();
        rf();int [] d = readArr(n);
        boolean [] f = new boolean[n + 1];
        ArrayList<Integer> a = new ArrayList<>();
        for(int i : d){
            if(i <= n && !f[i]){
                f[i] = true;
            }else{
                a.add(i);
            }
        }
        Collections.sort(a);
        int idx = 0;
        for(int i = 1; i <= n; i++){
            if(!f[i]){
                if((a.get(idx) - 1) / 2 < i){
                    addAns(-1);
                    return ;
                }
                idx++;
            }
        }
        addAns(a.size());
		
    }

    private void run() throws Exception {
		
        int T = 1;
		rf();
		T = ri();
		
		preparation();
        while (T-- > 0) {
            solve();
            if (T != 0) {
                clear();
            }
        }
        printAns();
    }

    public static void main(String[] args) throws Exception {
        new X1631().run();
    }

    StringBuilder sb = new StringBuilder();
    BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer strT;

	private void addAns(int a){
		sb.append(a + "\n");
	}
	
	private void addAns(long a){
		sb.append(a + "\n");
	}
	
	private void addAns(String s){
		sb.append(s + "\n");
	}

    private void rf() throws IOException {
        strT = new StringTokenizer(infile.readLine());
    }


    private int ri() {
        return Integer.parseInt(strT.nextToken());
    }

    private long rl() {
        return Long.parseLong(strT.nextToken());
    }

	private char [] rs2c(){
		return strT.nextToken().toCharArray();
	}
	
	private String rs(){
		return strT.nextToken();
	}

    private void printAns() {
        System.out.println(sb);
    }

    private int[] readArr(int N) throws Exception {
        int[] arr = new int[N];
        for (int i = 0; i < N; i++) {
            arr[i] = Integer.parseInt(strT.nextToken());
        }
        return arr;
    }

	private long[] readArr2(int N) throws Exception {
        long[] arr = new long[N];
        for (int i = 0; i < N; i++) {
            arr[i] = Long.parseLong(strT.nextToken());
        }
        return arr;
    }

	private void print(String format, Object ... args){
		System.out.println(String.format(format,args));
	}
	
	// for interactive problem. flush output after print.
	private void iprint(String format, Object ... args){
		System.out.println(String.format(format,args));
		System.out.flush();
	}

    private void print(int[] arr, int x) {
        //for debugging only
        for (int i = 0;i < min(arr.length, x); i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    }
    
    private void print(long[] arr, int x) {
        //for debugging only
        for (int i = 0;i < min(arr.length, x); i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    }

    
}
