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
/*in the name of Allah the most beneficent the most merciful.*/
#include<bits/stdc++.h>

#define ll long long
#define pb push_back
#define MAX 1e18
#define MIN -1e18
// #define MOD 1000000007
//#define mod2 20071027
#define MOD 998244353
#define base 139
// #define MOD 1000074259
#define base2 31
#define memz(a) memset(a, 0, sizeof(a))
#define memn(a) memset(a, -1, sizeof(a))
#define in1(a) scanf("%lld", &a)
#define in1i(a) scanf("%d", &a)
#define in2i(a, b) scanf("%d%d", &a, &b)
#define in2(a, b) scanf("%lld%lld", &a, &b)
#define TC(c) printf("Case #%lld: ", ++c)
#define out(x) cout << #x << " -> " << x << endl;
#define FAST ios_base::sync_with_stdio(false); cin.tie(NULL);
#define FILE   freopen("input.txt", "r", stdin);  freopen("out.txt", "w", stdout);
#define pii pair<ll, ll>
#define piii pair<ll, pair<ll, ll>>
#define PQ priority_queue
using namespace std;
const ll N = 510;
ll t, n, x, a[N], b, mem[510][510][2], cl;
ll dp(ll pos, ll s, ll sw) {
	ll &ret = mem[pos][s][sw];
	if (~ret) return ret;
	if (pos > n) {
		return 0;
	}
	ret = MAX;
	ll cs = s, pre = a[pos - 1];
	if (sw) {
		swap(cs, pre);
	}
	if (a[pos] > cs && cs>=pre) {
		ret = min(ret, 1 + dp(pos + 1, cs, 1));
	}
	if (a[pos] >= pre) {
		ret = min(ret, dp(pos + 1, cs, 0));
	}
	return ret;
}
int solve()
{
	in1(t);
	while (t--) {
		in2(n, x);
		memn(mem);
		for (int i = 1; i <= n; i++) in1(a[i]);
		ll ret = dp(1, x, 0);
		if (ret == MAX) ret = -1;
		printf("%lld\n", ret);
	}
	return 0;
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	return solve();
}
/*

*/
