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 ar array
#define ll long long

const int MAX_N = 1e5 + 1;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;



void solve() {
    int n, m; cin >> n >> m;
    vector<int> a(n);
    for (int &x : a) cin >> x;
    vector<int> sf(n + 1, 0);
	sf[n] = sf[n - 1] = 1;
	for (int i = n - 2; i >= 0; i--) 
		if (a[i] <= a[i + 1])
			sf[i] = sf[i + 1];
	
	int ans = 0, uk = 0;
	while (true) {
		int np = uk;
		while (np < n && a[np] <= m)
			np++;
		for (int i= uk; i < np; i++) {
			if (i == 0) continue;
			if (a[i - 1] > a[i]) {
				cout << -1 << "\n";
				return;
			}
		}
		if (sf[np])
			break;
		
		swap(a[np], m);
		ans++;
		uk = np;
	}
	cout << ans << "\n";
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    int tc = 1;
    cin >> tc;
    for (int t = 1; t <= tc; t++) {
        // cout << "Case #" << t  << ": ";
        solve();
    }
}