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;
const int mod =  1000000007;
const int LIM = 100005;
#define int long long
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define mem(a,b) memset(a,b,sizeof(a))
#define all(a) a.begin() , a.end()
#define pii pair<int,int>
#define vi vector<int>
#define endl '\n'
#define pb push_back
#define sp <<" "<<
#define ss second
#define ff first
int power(int x, int n){
    int res = 1;
    while(n){
          if(n&1){
                res = res * x % mod;
          }
          x = x * x % mod;
          n>>=1;
    }
    return (res%mod);
}
/*bool prime[1000006];
void sieve(int n)
{
      memset(prime, true, sizeof(prime));
      int rootn = (int)sqrt(n);
      for (int p = 2; p <= rootn; p++)
            if(prime[p] == true)
                  for(int i = p*p; i <= n; i += p)
                        prime[i] = false;
      prime[1] = 0;

}*/
/*
int fac[300005];
int ncr(int n,int r)
{
      return fac[n]*((power(fac[n-r], mod-2)*power(fac[r], mod-2))%mod)%mod;
}
*/
bool check(int a[],int n)
{
	for(int i=1;i<n;i++)
	{
		if(a[i]<a[i-1])return false;
	}
	return true;
}

int32_t main()
{
  IOS;
  int tt=1,n,x,k,y,z,i,j;
  cin>>tt;
  while(tt--){
    cin>>n>>x;
		int a[n];
		for(int i=0;i<n;i++)
		{
			cin>>a[i];
		}
		int ans=0;
		for(int i=0;i<=n;i++)
		{
			if(check(a,n))break;
			ans++;
			for(int j=0;j<n;j++)
			{
				if(a[j]>x)
				{
					swap(a[j],x);
					break;
				}
			}
		}
		if(check(a,n))cout<<ans<<endl;
		else cout<<-1<<endl;
  }
}
