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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main
{
    public static void main(String[] args)
    {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD
    {
        public void solve(int testNumber, Scanner in, PrintWriter out)
        {
            int T = in.nextInt();
            while (T-- > 0)
            {
                solveOne(in, out);
            }
        }

        private void solveOne(Scanner in, PrintWriter out)
        {
            int N = in.nextInt();
            int X = in.nextInt();
            int nums[] = CPUtils.readIntArray(N, in);

            int ops = 0;
            for (int i = 0; i < N && !isSorted(nums); i++)
            {
                if (nums[i] > X)
                {
                    int t = nums[i];
                    nums[i] = X;
                    X = t;
                    ops++;
                }
            }
            if (!isSorted(nums))
            {
                out.println(-1);
                return;
            }

            //System.out.println(Arrays.toString(nums));
            out.println(ops);
        }

        private boolean isSorted(int nums[])
        {
            for (int i = 0; i < nums.length - 1; i++)
            {
                if (nums[i] > nums[i + 1])
                {
                    return false;
                }
            }
            return true;
        }

    }

    static class CPUtils
    {
        public static int[] readIntArray(int size, Scanner in)
        {
            int[] array = new int[size];
            for (int i = 0; i < size; i++)
            {
                array[i] = in.nextInt();
            }
            return array;
        }

    }
}

