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      : Dinesh Verra
    College     : ABV-IIITM
    Date        : 06/12/2020
*/
// #pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
// #include <boost/multiprecision/cpp_int.hpp> 
// using namespace boost::multiprecision; 
using namespace std;
typedef int ll;
// #define ll int
typedef unsigned long long ull;
#define cu continue
#define br break
#define pb push_back
#define eb emplace_back
#define mod 1000000007
#define inf 100000
#define pll pair <ll,ll>
#define min_pq priority_queue<pll,vector <pll>, greater <pll> >  
#define ar array
#define F first
#define S second
#define var(n) vector<ar<ll,n>>
#define vll vector <ll>
#define vpll vector <pll>
#define dbg(n) cout<<#n<<' '<<n<<endl;
#define all(v) v.begin(),v.end()
#define nl cout<<'\n'

template <typename A1>
void prn(A1&& arg)
{
	cout<<arg<<'\n';
}
template <typename A1, typename... A>
void prn(A1&& arg, A&&... args)
{
	cout<<arg<<' ';
	prn(args...);
}

#define N 502
ll dp[N][N][N],a[N],n;

ll solve(ll pos, ll x, ll prev) {

	if(pos>=n)
		return 0;

	ll& res=dp[pos][x][prev];
	if(res!=-1) return res;

	res=5005;
	if(a[pos]>=prev)
	res=min(res,solve(pos+1,x,a[pos]));
	
	if(a[pos]>x && x>=prev)
	res=min(res,1+solve(pos+1,a[pos],x));

	return res;
}

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

	#ifndef ONLINE_JUDGE
	freopen("/home/dinesh_verra/cpp/input.txt","r",stdin);
	// freopen("/home/dinesh_verra/cpp/A.txt","w",stdout);
	#endif

	ll t;
	cin>>t;
	while(t--) {

		ll x;
		cin>>n>>x;
		for(ll i=0;i<n;i++) cin>>a[i];
		for(ll i=0;i<=n;i++)
			for(ll j=0;j<N;j++)
				for(ll k=0;k<N;k++) dp[i][j][k]=-1;
		ll ans=solve(0,x,0);
		if(ans>505) ans=-1;
		prn(ans);
	}	
}