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
/*
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
ordered_set -> less -> less_equal
ordered_set find_by_order(k) -> k element
ordered_set order_of_key(k) -> returns the number of elements in a set strictly smaller than k
*/
#include <bits/stdc++.h>
#include <random>
//#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
//using namespace __gnu_pbds;

//template<typename T> using ordered_set=tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

const long long maxn = 3e5 + 123;
const long long inf = 1e9 + 123;
const long long linf = 1e18 + 123;
const long long mod = 1e9 + 7;
const double eps = 1e-9;
const double pi = acos(-1);
int dx[8] = {0, 1, -1, 0, 1, -1, 1, -1};
int dy[8] = {1, 0, 0, -1, 1, 1, -1, -1};

int a[maxn];

bool ok[maxn];

int main(){
    //freopen("input.cpp", "r", stdin);
    //freopen("output.cpp", "w", stdout);
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);   
    int T;
    cin >> T;
    while (T --){
        int n, x;
        cin >> n >> x;
        int cntr = 0;
        int ans = 0;
        for (int i = 1; i <= n; i++){
            cin >> a[i];
            ok[i] = 0;
            if (a[i - 1] > a[i]){
                cntr++;
                ok[i] = 1;
            }
        }
        for (int i = 1; i <= n; i++){
            if (cntr >= 1 && a[i] > x){
                swap(a[i], x);
                ans ++;
            }
            if (ok[i] == 1){
                cntr--;
            }
        }
        bool OK = 0;
        for (int i = 1; i <= n; i++){
            if (a[i - 1] > a[i]){
                OK = 1;
            }
        }
        if (OK == 1){
            cout << "-1\n";
        }
        else{
            cout << ans << "\n";
        }
    }
    exit(0);
}
