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.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;

// Paprika and Permutation
public class PaprikaAndPermutation {

    public static void main(String[] args) throws IOException {
        PaprikaAndPermutation.solution();
    }

    private static void solution() throws IOException {
        PrintWriter printWriter = new PrintWriter(System.out);
        MyScanner scanner = new MyScanner();
        int numCases = scanner.nextInt();
        for (int i = 0; i < numCases; i++) {
            int n = scanner.nextInt();
            List<Integer> arr = new ArrayList<>();
            for (int j = 0; j < n; j++) {
                arr.add(scanner.nextInt());
            }
            Collections.sort(arr);
            boolean[] seen = new boolean[n + 1];
            int x = 1;
            int ans = 0;
            for (int j = 0; j < n; j++) {
                if (arr.get(j) >= 1 && arr.get(j) <= n && !seen[arr.get(j)]) {
                    seen[arr.get(j)] = true;
                } else {
                    if (x * 2 >= arr.get(j)) {
                        break;
                    } else {
                        seen[x] = true;
                        ans++;
                    }
                }
                while (x < seen.length && seen[x]) {
                    x++;
                }
            }
            if (x == n + 1) {
                printWriter.println(ans);
            } else {
                printWriter.println("-1");
            }
        }
        printWriter.close();
    }

    public static class MyScanner {
        private BufferedReader bufferedReader;
        private StringTokenizer st;

        public MyScanner() {
            bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bufferedReader.readLine());
            }
            return st.nextToken();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
    }
}
