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;
//#define int long long
#define f first
#define s second
const int mod=1e9+7;
int dp[501][501][501];
int dp2[501][501];
int a[501],n,x;
int fun(int idx,int last,int cur)
{
    if(idx==n) return 0;
    int &ret=dp[idx][last][cur];
    if(~ret) return ret;
    ret=1e9;
    if(a[idx]>=last)
        ret=fun(idx+1,a[idx],cur);
    if(cur>=last&&a[idx]>=cur)
        ret=min(ret,1+fun(idx+1,cur,a[idx]));
    return ret;
}
signed main()
{
    int t;
    cin>>t;
    while(t--)
    {

        scanf("%d%d",&n,&x);
          for(int i=0;i<=n;i++) for(int j=0;j<=500;j++) for(int k=0;k<=500;k++) dp[i][j][k]=-1;
        for(int i=0;i<n;i++)
            scanf("%d",a+i);
        int ans=fun(0,0,x);
        if(ans>=1e9) puts("-1");
        else
        printf("%d\n",ans);
    }

   return 0;
}
