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>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize ("-ffloat-store") // to restrict undesirable precision
#pragma GCC optimize ("-fno-defer-pop")// to pop argument of function as soon as it returns
#define all(a) a.begin(),a.end()
#define ll long long int
#define ld long double
ll power(ll a,ll b,ll m){ if(b==0) return 1; if(b==1) return a%m; ll t=power(a,b/2,m)%m; t=(t*t)%m; if(b&1) t=((t%m)*(a%m))%m; return t;}
ll modInverse(ll a, ll m) { return power(a, m-2, m); }
#define ps push_back
#define fs first
#define sc second
#define takeline cin.ignore();
#define iactive cout.flush();
#define N 505
#define endl "\n"
#define mod 1000000007
//((1.0l)*BIG MULTIPLY MAGIC?)
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//-----------------------------------------------
ll n,x,ar[N],dp[N][N];
ll minval(ll i,ll cur_x){
	if(i==n) return 0;
	
	if(dp[i][cur_x]!=-1){
		return dp[i][cur_x];
	}
	if(ar[i]<=ar[i+1]){
		ll a,b=1e18;
		a=minval(i+1,cur_x);
		if(ar[i]>cur_x&&(i==1||cur_x>=ar[i-1])) b=minval(i+1,ar[i])+1;
		return dp[i][cur_x]=min(a,b);
	}
	else{
		if(ar[i]<=cur_x||cur_x>ar[i+1]||(i>1&&cur_x<ar[i-1])) return dp[i][cur_x]=1e18;
		else return dp[i][cur_x]=minval(i+1,ar[i])+1;
	}
}
int main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	
	int t;
	cin>>t;
	while(t--){
		ll i,j,k,l;
		cin>>n>>x;
		for(i=1;i<=n;i++){
			cin>>ar[i];
		}
		memset(dp,-1,sizeof(dp));
		ll an=minval(1,x);
		if(an==1e18) cout<<"-1"<<endl;
		else cout<<an<<endl;
	}
	return 0;
}
