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>
#pragma GCC target ("avx,avx2,fma")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#define X first
#define Y second
#define debug(x) cerr << #x << ": " << x << endl;
#define endl '\n'
#define pb push_back
#define rep(i,l,r) for (int i=l; i<r; i++)
#define repr(i,r,l) for (int i=r; i>=l; i--)
using namespace std;
typedef long long  int ll;
typedef pair<int,int> pll;
const long long int N=5e2+30,mod = 1e9+7,inf=2e5;
const long double eps = 0.0001;
int a[N];

int main(){
    int T;
    cin >> T;
    while (T--){
        int n,x,j=-1;
        bool f = 0;
        cin >> n >> x;
        rep(i,0,n){
            cin >> a[i];
            if (i && a[i] < a[i-1]) j = i-1;
        }
        int ans = 0;
        rep(i,0,n){
            if (!i){
                if (j > 0 && a[i] > x){
                    swap(a[i],x);
                    ans++;
                }
                continue;
            }
            if ((a[i] < a[i-1] || i <= j) && a[i-1] > x){
                swap(a[i-1],x);
                ans++;
            }
            if (a[i] < a[i-1]){
                f = 1;
                break;
            }
        }
        if (f){
            cout << -1 << endl;
            continue;
        }
        //if (n >= 2 && a[0] > a[1]) ans++;
        cout << ans << endl;
    }
    return 0;
}

