"It does not matter how slowly you go as long as you do not stop." - Confucius
You are given an array $A_1, A_2, \ldots, A_N$ and an integer $K$. For each subarray $S = [A_l, A_{l+1}, \ldots, A_r]$ ($1 \le l \le r \le N$):
- Let's define an array $B$ as $S$ concatenated with itself $m$ times, where $m$ is the smallest integer such that $m(r-l+1) \ge K$.
- Next, let's sort $B$ and define $X = B_K$, i.e. as a $K$-th smallest element of $B$. Note that $|B| \ge K$.
- Then, let's define $F$ as the number of occurrences of $X$ in $S$.
- The subarray $S$ is beautiful if $F$ occurs in $S$ at least once.
Find the number of beautiful subarrays of $A$. Two subarrays $A_l, A_{l+1}, \ldots, A_r$ and $A_p, A_{p+1}, \ldots, A_q$ are different if $l \neq p$ or $r \neq q$.

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first line of each test case contains two space-separated integers $N$ and $K$. 
- The second line contains $N$ space-separated integers $A_1, A_2, \ldots, A_N$.

-----Output-----
For each test case, print a single line containing one integer - the number of beautiful subarrays.

-----Constraints-----
- $1 \le T \le 5$
- $1 \le N \le 2,000$
- $1 \le K \le 10^9$
- $1 \le A_i \le 2000$ for each valid $i$

-----Subtasks-----
Subtask #1 (20 points): $1 \le N \le 200$
Subtask #2 (80 points): original constraints

-----Example Input-----
1
3 3
1 2 3

-----Example Output-----
3

-----Explanation-----
Example case 1: There are six subarrays of $A$: $[1]$, $[2]$, $[3]$, $[1, 2]$, $[2, 3]$, $[1, 2, 3]$. The corresponding arrays $B$ are $[1, 1, 1]$, $[2, 2, 2]$, $[3, 3, 3]$, $[1, 2, 1, 2]$, $[2, 3, 2, 3]$, $[1, 2, 3]$.
Three of the subarrays are beautiful: $[1]$, $[1, 2]$ and $[1, 2, 3]$. For these subarrays, $X$ is $1$, $2$ and $3$ respectively (for example, for $S = [1, 2]$, $B = [1, 2, 1, 2]$ is sorted to $[1, 1, 2, 2]$ and $X = 2$ is the $3$-rd element). Then, $F = 1$ for each of these subarrays, and each of these subarrays contains $1$.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import heapq\n\nt=int(input())\nfor y in range(t):\n (n, k )=(int(x) for x in input().split())\n places=[]\n for l in range(1,n+1):\n  m=(k-1)//l+1\n  shoot=m*l-k\n  place=shoot//m+1\n  places.append(place)\n A=[int(x) for x in input().split()]\n heaps=[[-x] for x in A]\n assert(len(A)==n)\n ans=A.count(1)\n for l in range(1,n):\n  elements=[0]*2001\n  for e in A[:l+1]:\n   try:\n    elements[e]+=1\n   except:\n    elements[e]=1\n  for s in range(0,n-l):\n   heapq.heappush(heaps[s],-A[s+l]) \n   if places[l]>2:\n    reserve=[]\n    for i in range(places[l]):\n     reserve.append(heapq.heappop( heaps[s]))\n    for r in reserve:\n     heapq.heappush(heaps[s],r)\n    x=-reserve[-1]\n   elif (places[l]==2):\n    x=max(-heaps[s][1],heaps[s][2])\n   else:\n    x=-heaps[s][0]\n   f=elements[x]\n   try:\n    if elements[f]>0:\n     ans+=1\n   except:\n    pass\n   if s<n-l-1:\n    try:\n     elements[A[s]]-=1\n     elements[A[s+l+1]]+=1\n    except: \n     elements[A[s+l+1]]=1\n print(ans)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 38683