You are given a sequence a consisting of n integers a_1, a_2, ..., a_n, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a_1 ≤ a_2 ≤ a_3 ≤ ... ≤ a_n holds).

To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1 ≤ i ≤ n and a_i > x, and swap the values of a_i and x.

For example, if a = [0, 2, 3, 5, 4], x = 1, the following sequence of operations is possible:

  1. choose i = 2 (it is possible since a_2 > x), then a = [0, 1, 3, 5, 4], x = 2; 
  2. choose i = 3 (it is possible since a_3 > x), then a = [0, 1, 2, 5, 4], x = 3; 
  3. choose i = 4 (it is possible since a_4 > x), then a = [0, 1, 2, 3, 4], x = 5. 



Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.

Input

The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.

Each test case consists of two lines. The first line contains two integers n and x (1 ≤ n ≤ 500, 0 ≤ x ≤ 500) — the number of elements in the sequence and the initial value of x.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 500).

The sum of values of n over all test cases in the input does not exceed 500.

Output

For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or -1, if it is impossible.

Example

Input


6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324


Output


3
0
0
-1
1
3
import java.io.PrintStream;
import java.util.Scanner;

public class TaskD {

    public static void main(final String[] args) {
        final Scanner in = new Scanner(System.in);
        final PrintStream out = System.out;

        final int t = in.nextInt();

        for (int i = 0; i < t; i++) {
            final int n = in.nextInt();
            int x = in.nextInt();

            final int[] a = new int[n];

            for (int j = 0; j < n; j++) {
                a[j] = in.nextInt();
            }

            int ops = 0;

            for (int j = 0; j < n && !isSorted(a); j++) {
                if (a[j] > x) {
                    final int temp = a[j];
                    a[j] = x;
                    x = temp;
                    ops++;
                }
            }

            out.println(isSorted(a) ? ops : -1);
        }

    }

    private static boolean isSorted(final int[] array) {
        for (int i = 1; i < array.length; i++) {
            if (array[i] < array[i - 1]) {
                return false;
            }
        }

        return true;
    }

}
