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 mod 1000000007
#define N 505
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
template<typename T>
inline void in(T &x)
{
	x=0;
	char c=getchar();
	bool fu=0;
	while(c<'0'||c>'9'){if(c=='-')fu=1;c=getchar();}
	while(c>='0'&&c<='9')x=(x<<1)+(x<<3)+(c^48),c=getchar();
	x=fu?-x:x;
}
int a[N],n,x;
void work()
{
	int ans=0;
	in(n);in(x);
	for(int i=0;i<n;++i)in(a[i]);
	for(int i=0;i<n;++i)
	{
		if(i>0)
		{
			if(a[i]<a[i-1])
			{
				if(x>a[i]){puts("-1");return;}
				for(int j=0;j<i;++j)
				{
					if(a[j]>x)swap(a[j],x),++ans;
				}
				if(a[i-1]>a[i]){puts("-1");return;}
			}
		}
	}
	printf("%d\n",ans);
}
int main()
{
	int T;
	in(T);
	while(T--)work();
	return 0;
}