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;
typedef long long ll; 
const int mod = 998244353, mxn = 1e5 + 1;

int solve(){
	int n, x;
	cin>> n>> x;
	int a[n];
	for(int i = 0; i < n; i++)
		cin>> a[i];
	int ans = 0;
	while(true){
		int ptr = -1;
		for(int i = 1; i < n; i++){
			if(a[i] < a[i - 1]){
				ptr = i;
				break;
			}
		}
		if(ptr == -1)
			break;
		ptr = -1;
		for(int i = 0; i < n; i++){
			if(a[i] > x){
				ptr = i;
				break;
			}
		}
		if(ptr == -1){
			return cout<< "-1\n", 0;
		}
		int tmp = a[ptr];
		a[ptr] = x;
		x = tmp;
		ans++;
	}
	cout<< ans<< "\n";
	return 0;
}

int main(){
	ios_base::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL);
	int t = 1;
	cin>> t;
	while(t--) 
		solve();
	return 0;
}