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 fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int long long
#define FIN(i,n) for(int i=0;i<n;i++)
#define loop(i, a, b) for (int i = a; i < b; i++)
#define vp vector<pair < int, int > >
#define vll vector<long long>
#define f first
#define s second
#define MINV(v) *min_element(v.begin(),v.end())
#define MINIV(v) min_element(v.begin(),v.end())-v.begin()
#define all(v) v.begin(), v.end()
#define pb push_back
#define TR(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)

// make_pair()
//vector<vector<char>> v(r, vector<char> (c,0));
int mod=1e9+7;
int INF=1e18;

bool isPrime(int n){if(n<=1)return false;if(n<=3)return true;
    if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)
    if(n%i==0||n%(i+2)==0)return false;return true;}

int nCr(int n,int r){int res=1;if(n<r)return -1;r=min(r,n-r);
    loop(i,0,r){res*=(n-i);res/=(i+1);}return res;}

int nPr(int n,int r){int res=1;if(n<r)return -1;
    loop(i,0,r)res*=(n-i);return res;}

void solve()
{
    int n,x;
    cin>>n>>x;
    vll v(n);
    FIN(i,n){ cin>>v[i]; }
    vll v2;
    v2=v;
    sort(all(v2));
    if(v2==v){
        cout<<0<<"\n";return;
    }
    int ct=0;
    for(int i=0;i<n;i++){
        if(v[i]>x){
            swap(x,v[i]);
            ct++;
            v2.clear();
            v2=v;
            sort(all(v2));
            if(v2==v){
                cout<<ct<<"\n";
                return;
            }
        }
    }
    cout<<-1<<"\n";
}

signed main(){
    fast
    int t=1;
    cin>>t;
    while(t--)
        solve();
    return 0;
}