description: The new generation external memory contains an array of integers a[1 … n] = [a_1, a_2, …, a_n].
  
  This type of memory does not support changing the value of an arbitrary element. Instead, it allows you to cut out any segment of the given array, cyclically shift (rotate) it by any offset and insert it back into the same place.
  
  Technically, each cyclic shift consists of two consecutive actions: 
  
    1. You may select arbitrary indices l and r (1 ≤ l < r ≤ n) as the boundaries of the segment. 
    2. Then you replace the segment a[l … r] with it's cyclic shift to the left by an arbitrary offset d. The concept of a cyclic shift can be also explained by following relations: the sequence [1, 4, 1, 3] is a cyclic shift of the sequence [3, 1, 4, 1] to the left by the offset 1 and the sequence [4, 1, 3, 1] is a cyclic shift of the sequence [3, 1, 4, 1] to the left by the offset 2. 
  
  
  
  For example, if a = [1, \color{blue}{3, 2, 8}, 5], then choosing l = 2, r = 4 and d = 2 yields a segment a[2 … 4] = [3, 2, 8]. This segment is then shifted by the offset d = 2 to the left, and you get a segment [8, 3, 2] which then takes the place of of the original elements of the segment. In the end you get a = [1, \color{blue}{8, 3, 2}, 5].
  
  Sort the given array a using no more than n cyclic shifts of any of its segments. Note that you don't need to minimize the number of cyclic shifts. Any method that requires n or less cyclic shifts will be accepted.
  
  Input
  
  The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.
  
  The next 2t lines contain the descriptions of the test cases. 
  
  The first line of each test case description contains an integer n (2 ≤ n ≤ 50) — the length of the array. The second line consists of space-separated elements of the array a_i (-10^9 ≤ a_i ≤ 10^9). Elements of array a may repeat and don't have to be unique.
  
  Output
  
  Print t answers to all input test cases. 
  
  The first line of the answer of each test case should contain an integer k (0 ≤ k ≤ n) — the number of actions to sort the array. The next k lines should contain descriptions of the actions formatted as "l r d" (without quotes) where l and r (1 ≤ l < r ≤ n) are the boundaries of the segment being shifted, while d (1 ≤ d ≤ r - l) is the offset value. Please remember that only the cyclic shifts to the left are considered so the chosen segment will be shifted by the offset d to the to the left.
  
  Note that you are not required to find the minimum number of cyclic shifts needed for sorting. Any sorting method where the number of shifts does not exceed n will be accepted.
  
  If the given array a is already sorted, one of the possible answers is k = 0 and an empty sequence of cyclic shifts.
  
  If there are several possible answers, you may print any of them.
  
  Example
  
  Input
  
  
  4
  2
  2 1
  3
  1 2 1
  4
  2 4 1 3
  5
  2 5 1 4 3
  
  
  Output
  
  
  1
  1 2 1
  1
  1 3 2
  3
  2 4 1
  2 3 1
  1 3 2
  4
  2 4 2
  1 5 3
  1 2 1
  1 3 1
  
  Note
  
  Explanation of the fourth data set in the example: 
  
    1. The segment a[2 … 4] is selected and is shifted to the left by 2: [2, \color{blue}{5, 1, 4}, 3] \longrightarrow [2, \color{blue}{4, 5, 1}, 3] 
    2. The segment a[1 … 5] is then selected and is shifted to the left by 3: [\color{blue}{2, 4, 5, 1, 3}] \longrightarrow [\color{blue}{1, 3, 2, 4, 5}] 
    3. After that the segment a[1 … 2] is selected and is shifted to the left by 1: [\color{blue}{1, 3}, 2, 4, 5] \longrightarrow [\color{blue}{3, 1}, 2, 4, 5] 
    4. And in the end the segment a[1 … 3] is selected and is shifted to the left by 1: [\color{blue}{3, 1, 2}, 4, 5] \longrightarrow [\color{blue}{1, 2, 3}, 4, 5] 
solution: 
  # Author : raj1307 - Raj Singh
  # Date   : 28.09.2021
  
  from __future__ import division, print_function
  
  import os,sys
  from io import BytesIO, IOBase
  
  if sys.version_info[0] < 3:
      from __builtin__ import xrange as range
      from future_builtins import ascii, filter, hex, map, oct, zip
  
  
  def ii():  return int(input())
  def si():  return input()
  def mi():  return map(int,input().strip().split(" "))
  def msi(): return map(str,input().strip().split(" "))
  def li():  return list(mi())
  
  def dmain():
      sys.setrecursionlimit(1000000)
      threading.stack_size(1024000)
      thread = threading.Thread(target=main)
      thread.start()
      
  #from collections import deque, Counter, OrderedDict,defaultdict
  #from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
  #from math import log,sqrt,factorial,cos,tan,sin,radians
  #from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
  #from decimal import *
  #import threading
  #from itertools import permutations
  #Copy 2D list  m = [x[:] for x in mark] .. Avoid Using Deepcopy
  
  abc='abcdefghijklmnopqrstuvwxyz'
  abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
  mod=1000000007
  #mod=998244353
  inf = float("inf")
  vow=['a','e','i','o','u']
  dx,dy=[-1,1,0,0,1,-1,1,-1],[0,0,1,-1,1,-1,-1,1]
  def getKey(item): return item[1] 
  def sort2(l):return sorted(l, key=getKey,reverse=True)
  def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
  def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
  def decimalToBinary(n): return bin(n).replace("0b","")
  def ntl(n):return [int(i) for i in str(n)]
  def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
  
  def ceil(x,y):
      if x%y==0:
          return x//y
      else:
          return x//y+1
  
  def powerMod(x,y,p):
      res = 1
      x %= p
      while y > 0:
          if y&1:
              res = (res*x)%p
          y = y>>1
          x = (x*x)%p
      return res
  
  def gcd(x, y):
      while y:
          x, y = y, x % y
      return x
      
  def isPrime(n) : # Check Prime Number or not 
      if (n <= 1) : return False
      if (n <= 3) : return True
      if (n % 2 == 0 or n % 3 == 0) : return False
      i = 5
      while(i * i <= n) : 
          if (n % i == 0 or n % (i + 2) == 0) : 
              return False
          i = i + 6
      return True
  
  
  
  def read():
      sys.stdin  = open('input.txt', 'r')  
      sys.stdout = open('output.txt', 'w') 
  
  
  
  
  def main():
  
  
      
      for _ in range(ii()):
  
  
          n=ii()
          a=li()
          b=a[:]
          b.sort()
          ans=[]
  
  
          for i in range(n):
              p=-1
              for j in range(i,n):
                  if a[j]==b[i]:
                      p=j
                      break
  
              a[i+1:p+1]=a[i:p]
              a[i]=a[j]
              if i!=p:
                  ans.append([i+1,p+1,p-i])
  
          print(len(ans))
          for i in range(len(ans)):
              print(*ans[i])
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
      
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  # region fastio
  # template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
  
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  def print(*args, **kwargs):
      """Prints the values to a stream, or to sys.stdout by default."""
      sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
      at_start = True
      for x in args:
          if not at_start:
              file.write(sep)
          file.write(str(x))
          at_start = False
      file.write(kwargs.pop("end", "\n"))
      if kwargs.pop("flush", False):
          file.flush()
  
  
  if sys.version_info[0] < 3:
      sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
  else:
      sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  # endregion
  
  
  if __name__ == "__main__":
      #read()
      main()
      #dmain()
  
  # Comment Read()
  