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 forn(i,n) for(int i=0;i<(int)(n);i++)
#define si(c) ((int)(c).size())
#define forsn(i,s,n) for(int i = (int)(s); i<((int)n); i++)
#define dforsn(i,s,n) for(int i = (int)(n)-1; i>=((int)s); i--)
#define all(c) (c).begin(), (c).end()
#define D(a) cerr << #a << "=" << (a) << endl;
#define pb push_back
#define eb emplace_back
#define mp make_pair
 
typedef long long int ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());


int go(vi a,  vi b, int x) {
    sort(all(b));
    int n = si(a), put = x, ans = 0;
    bool ok = 1;
    forn(i,n) {
        if (b[i] != a[i]) {
            ok &= b[i] < a[i];
            ok &= b[i] == put;
            put = a[i];
            ans++;
        }
    }
    if (!ok) return n+2;
    return ans;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);

    int T; cin >> T;
    while (T--) {
        int n,x; cin >> n >> x;
        vi a(n);
        forn(i,n) cin >> a[i];

        int ans = go(a,a,x);
        forn(i,n) {
            auto b = a;
            b[i] = x;
            ans = min(ans, go(a,b,x));
        }
        if (ans > n) ans = -1;
        cout << ans << '\n';
    }

    return 0;
}
