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 Mod                  1000000007
#define PI                   3.14159265358979323846
#define ll                   long long
#define lld                  long double
#define ld                   double
#define Vi					 vector<int>
#define Vll					 vector<long long>
#define Vc					 vector<char>

#define lcm(m, n)            ((m/__gcd(m,n))*n)
#define deb(x)				 cout << #x << " " << x << endl // used to debug any variable
#define scn(str)             scanf("%s",str)
#define pri(str)             printf("%s\n",str)

#define rfo(i, n) 			 for(int i = n-1; i >=0; i--)
#define fo(i, n)             for(int i = 0; i < n; i++)
#define ifo(i, k, n)         for(int i = k; i < n; i++)
#define inp(arr, n)          for(int ele=0; ele<n; ele++) cin >> arr[ele];
#define ms(arr, v)           memset(arr, v, sizeof(arr)) // memset sets the elements to given index (here all: sizeofarr) of arr[] to value v
#define all(v)               v.begin(), v.end()

#define fastIO                 ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);

void solve();

int main() {
	fastIO

	// Forget ratings, enjoy the experience

	// clock_t time_req; 
    // time_req = clock();
	
	int t = 1;
    int i = 1;
    // Vll v;
    // v.push_back(1);
    // while (v.back() < 1000000) {
    //     i++;
    //     v.push_back(v.back() + i);
    // }
	cin >> t;
	while(t--) {
		solve();
	}

	// time_req = clock()- time_req; 
	// cout << "Processor time taken for multiplication: " << (float)time_req/CLOCKS_PER_SEC << " seconds" << endl;
	
	return 0;
}
// write code here
void solve() {
    ll x, n, c = 0, j;
    cin >> n >> x;
    j = n - 1;
    Vll a(n);
    inp(a, n);
    for (int k = n - 1; k > 0; k--) {
        if (a[k] >= a[k - 1]) {
            j = k - 1;
        }
        else break;
    }
    // deb(j);
    fo(i, n - 1) {
        if (i < j && a[i] > x) {
            swap(a[i], x);
            c++;
            i--;
            continue;
        }
        if (a[i] > a[i + 1]) {
            cout << "-1\n";
            return;
        }
    }
    cout << c << "\n";
}

