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
/*#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse2")
/*#pragma GCC target("avx,avx2,fma")
#pragma GCC target ("sse4")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")*/
#include "bits/stdc++.h"
using namespace std;
#define ll int 
#define ld double 
#define rep(i,a,b) for(ll i=a; i<b; ++i)
#define BOOST std::ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define pb push_back
#define all(v) v.begin(),v.end()
#define sl(a) a.size()
#define mod 1000000007 
#define lcm(a,b) (a*b)/__gcd(a,b)
#define f first
#define s second
#define pi 2*acos(0.0)
bool isPowerOfTwo (ll x) {  return x && (!(x&(x-1)));  }
//╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋

//██████╗ ██████╗  █████╗ ███╗   ██╗███████╗██╗  ██╗██╗   ██╗
//██╔══██╗██╔══██╗██╔══██╗████╗  ██║██╔════╝██║  ██║██║   ██║
//██████╔╝██████╔╝███████║██╔██╗ ██║███████╗███████║██║   ██║
//██╔═══╝ ██╔══██╗██╔══██║██║╚██╗██║╚════██║██╔══██║██║   ██║
//██║     ██║  ██║██║  ██║██║ ╚████║███████║██║  ██║╚██████╔╝
//╚═╝     ╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═══╝╚══════╝╚═╝  ╚═╝ ╚═════╝ 
//╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋╋                                                        
                                                           
/********************************************************************************************/
ll cache[501][501][501];
vector<ll> arr(501);
ll n;
ll dp(ll indx, ll x, ll pre_val)
{
    if(indx>n) return 0;
    ll &ans=cache[indx][x][pre_val];
    if(ans!=-1) return ans;
    ans=1e9;
    if(arr[indx]>=pre_val) ans=min(ans,dp(indx+1,x,arr[indx]));
    if(arr[indx]>x and x>=pre_val) ans=min(ans,1+dp(indx+1,arr[indx],x));
    return ans;

}
void solve()
{
    ll x; cin>>n>>x;
    for(ll i=1; i<=n; ++i) cin>>arr[i];
    for(ll i=1; i<=n; ++i)
    {
        for(ll j=0; j<501; ++j)
        {
            for(ll k=0; k<501; ++k) cache[i][j][k]=-1;
        }
    }
    ll ans=dp(1,x,0);
    cout<<(ans==1e9?-1:ans)<<"\n";
}
int32_t main() 
{
    BOOST
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
    #endif
    int T=1; cin>>T;
    while(T--) solve();
    return 0;
}