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
/*Hatsune Miku 4ever!*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define _for(i,a,b) for(int i = (a);i < b;i ++)
#define _rep(i,a,b) for(int i = (a);i > b;i --)
#define INF 0x3f3f3f3f
#define lowbit(x) ((x)&(-x))
#define pb push_back
#define MIKU 39
#define Design ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define debug() printf("Miku Check OK!\n")
#define bigmiku 3939393939
#define maxn 539

int n, nx;
int aa[maxn], a[maxn];
int main()
{
	int _;
	scanf("%d",&_);
	while(_--)
	{
		scanf("%d%d",&n,&nx);
		_for(i,1,n+1)
			scanf("%d",&aa[i]);
		int fg = 1;
		_for(i,2,n+1)
		if(aa[i]<aa[i-1])
			fg = 0;
		if(fg)
		{
			printf("0\n");
			continue;
		}
		int ans = INF;
		_for(i,1,n+1)
		{
			int x = nx;
			_for(j,1,n+1)
				a[j] = aa[j];
			vector<int> b;
			_for(j,1,n+1)
				if(i==j)
					b.pb(x);
				else
					b.pb(a[j]);
			sort(b.begin(),b.end());
			int times = 0;
//			_for(j,0,n)
//				printf("%d ",b[j]);
//			cout << endl;
			_for(j,0,b.size())
			{
				if(b[j]==a[j+1])
					continue;
				if(a[j+1] <= x || a[j]>x)
					times = INF;
				else
				{
					times ++;
					swap(a[j+1],x);
//					_for(j,1,n+1)
//						printf("%d ",a[j]);
//					cout << endl;
				}
			}
			ans = min(ans,times);
		}
		printf("%d\n",ans==INF?-1:ans);
	}
	return 0;
}
