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
/*
author: Saftap
https://codeforces.com/profile/Saftap
*/
#define pb push_back
#define ll long long
#include <bits/stdc++.h>

using namespace std;

#define endl '\n'
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define minim(x) min_element((x).begin(), (x).end())
#define maxim(x) max_element((x).begin(), (x).end())
#define fori(i, n) for (long long (i) = 0; (i) < (n); ++(i))

template <class T>
istream& operator >>(istream &in, vector<T> &arr)
{
    for (T &i : arr)
    {
        in >> i;
    }
    return in;
}

template <class T>
ostream& operator <<(ostream &in, vector<T> &arr)
{
    for (T &i : arr)
    {
        in << i << " ";
    }
    return in;
}

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<vector<ll>, null_type, less<vector<ll>>, rb_tree_tag, tree_order_statistics_node_update> treap;
/*.order_of_key()
*.find_by_order()  iter*/



/*#include <ext/rope>
using namespace __gnu_cxx;
// .mutable_reference_at(i) =
// erase default(vector) + mbegin, mend
// insert default or rope
// .substr(ind, len)
#define mbegin mutable_begin
#define mend mutable_end
#define changei mutable_reference_at
template <class T>
ostream& operator <<(ostream &in, rope<T> &arr)
{
    for (const T &i : arr)
    {
        in << i << " ";
    }
    return in;
}*/



signed main()
{

    cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(false);
    //cout.precision(40);

    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);



    ll t;
    cin >> t;
    fori(_, t)
    {
        ll n, x;
        cin >> n >> x;
        vector<ll> m(n);
        cin >> m;

        vector<ll> jj;

        for (auto u: m)
        {
            jj.pb(u);
        }
        jj.pb(x);

        sort(all(jj));
        ll ans = 1e18;

        for (int i = 0; i < n + 1; ++i)
        {
            vector<ll> res;
            for (int j = 0; j < n + 1; ++j)
            {
                if (j == i)
                    continue;

                res.pb(jj[j]);
            }
            ll xx = x;
            ll fini = jj[i];
            ll a = 0;

            vector<ll> mm = m;
            for (int k = 0; k < n; ++k)
            {
                if (res[k] == mm[k])
                    continue;

                if (mm[k] <= xx)
                {
                    a = 1e18;
                    break;
                }
                else
                {
                    a += 1;
                    swap(xx, mm[k]);
                }
            }
            ans = min(ans, a);
        }
        if (ans == 1e18)
            ans = -1;
        cout << ans << endl;







    }

















}








