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 ii pair<int,int>
#define fi first
#define se second
#define ll long long 
#define all(x) x.begin(),x.end()
#define sz(x) (int)x.size()

const int mxn = 1e5;

void solve(){
	int n,x;
	cin>>n>>x;
	vector<int>a(n);
	for(int& i:a)cin>>i;
	int ans = n+1;
	if(is_sorted(all(a)))ans=0;
	for(int i=0; i<n; i++){
		vector<int>b(a);
		b[i]=x;
		sort(all(b));
		bool ok=true;
		int cnt=0;
		int swp = x;
		for(int j=0; j<n; j++){
			if(a[j]!=b[j]){
				cnt++;
				if(b[j]==swp&&a[j]>swp)swp = a[j];
				else ok=false;
			}
		}
		if(ok)ans=min(ans,cnt);
	}
	if(ans==n+1)ans=-1;
	cout<<ans<<"\n";
}

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int t;
	cin>>t;
	while(t--)solve();
	return 0;
}
