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
#include <bits/stdc++.h>
#define db(x) cout << (x) << '\n';
#define dbg(x) cerr << #x ": " << (x)<< '\n';
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vector<int> > adyList;

#define x first
#define y second
#define pb push_back
#define sqr(a) ((a) * (a))
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define forn(i, n) for(int i = 0; i < int(n); i++)
#define fore(i, l, r) for(int i = int(l); i < int(r); i++)
#define MAXN 107

const int INF = int(1e9);
const ll INF64 = ll(1e18);
const int MOD = int(1e9) + 7;
const ld EPS = 1e-9;
const ld PI = acos(-1.0);


int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        cin>>n>>m;
        vi v(n);
        for(auto &x : v)
            cin>>x;
        bool ok = 1;
        forn(i,n-1)
        {
            if(v[i] > v[i+1])
                ok = 0;
        }
        if(ok)
        {
            db(0)
            continue;
        }
        int c = 0;
        forn(i,n)
        {
            if(v[i] > m)
            {
                swap(v[i],m);
                c++;
                ok = 1;
                forn(j,n-1)
                {
                    if(v[j] > v[j+1])
                        ok = 0;
                }
                if(ok)
                {
                    db(c);
                    break;
                }
            }
        }
        if(ok)
            continue;
        ok = 1;
        forn(i,n-1)
        {
            if(v[i] > v[i+1])
                ok = 0;
        }
        if(ok)
        {
            db(c)
            continue;
        }
        else
            db(-1)

    }

    return 0;
}
