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;
#define INF 1000
const short N = 505;

short tc, cs = 1;
short n, x, arr[N], dp[N][N][N];
short vis[N][N][N];

short fun(short pos, short prevmx, short curx) {
  if (pos > n) return 0;
  short &ret = dp[pos][prevmx][curx];
  if (vis[pos][prevmx][curx] == cs) return ret;
  vis[pos][prevmx][curx] = 1;
  ret = INF;
  if (arr[pos] >= prevmx) {
    if (arr[pos] > curx) {
      if (curx >= prevmx) {
        short v = fun(pos + 1, curx, arr[pos]) + 1;
        if (v <= ret) ret = v;
      }
      short v = fun(pos + 1, arr[pos], curx);
      if (v <= ret) ret = v;
    } else {
      short v = fun(pos + 1, arr[pos], curx);
      if (v <= ret) ret = v;
    }
  }
  return ret;
}

void test_cases() {
  cin >> n >> x;
  for (int i = 1; i <= n; ++i) cin >> arr[i];
  short ans = fun(1, 0, x);
  if (ans >= INF) cout << "-1" << '\n';
  else cout << ans << '\n';
  cs++;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> tc;
  while (tc--) {
    test_cases();
  }
}