output_description: Print the number of pairs of integers i, j (1 \leq i < j \leq N) such that s_i
  is an anagram of s_j.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      s_1
      s_2
      :
      s_N
attempt: import numpy as np
  
  N = int(input())
  a = []
  b = []
  c = np.array([[0] * 26, [0] * 26, [0] * 26])
  count = 0
  for i in range(N):
      a.append(input())
  
  for i in range(N):
      b = a[i]
      for j in range(10):
          if b[j] == "a":
              c[i][0] += 1
          elif b[j] == "b":
              c[i][1] += 1
          elif b[j] == "c":
              c[i][2] += 1
          elif b[j] == "d":
              c[i][3] += 1
          elif b[j] == "e":
              c[i][4] += 1
          elif b[j] == "f":
              c[i][5] += 1
          elif b[j] == "g":
              c[i][6] += 1
          elif b[j] == "h":
              c[i][7] += 1
          elif b[j] == "i":
              c[i][8] += 1
          elif b[j] == "j":
              c[i][9] += 1
          elif b[j] == "k":
              c[i][10] += 1
          elif b[j] == "l":
              c[i][11] += 1
          elif b[j] == "m":
              c[i][12] += 1
          elif b[j] == "n":
              c[i][13] += 1
          elif b[j] == "o":
              c[i][14] += 1
          elif b[j] == "p":
              c[i][15] += 1
          elif b[j] == "q":
              c[i][16] += 1
          elif b[j] == "r":
              c[i][17] += 1
          elif b[j] == "s":
              c[i][18] += 1
          elif b[j] == "t":
              c[i][19] += 1
          elif b[j] == "u":
              c[i][20] += 1
          elif b[j] == "v":
              c[i][21] += 1
          elif b[j] == "w":
              c[i][22] += 1
          elif b[j] == "x":
              c[i][23] += 1
          elif b[j] == "w":
              c[i][24] += 1
          elif b[j] == "z":
              c[i][25] += 1
  for i in range(N):
      for j in range(1, N):
          if i != j:
              if (c[i] == c[j]).all():
                  count += 1
  
  print(count)
  
problem_description: Statement
  
  We will call a string obtained by arranging the characters contained in a
  string a in some order, an _anagram_ of a.
  
  For example, `greenbin` is an anagram of `beginner`. As seen here, when the
  same character occurs multiple times, that character must be used that number
  of times.
  
  Given are N strings s_1, s_2, \ldots, s_N. Each of these strings has a length
  of 10 and consists of lowercase English characters. Additionally, all of these
  strings are distinct. Find the number of pairs of integers i, j (1 \leq i < j
  \leq N) such that s_i is an anagram of s_j.
samples: [{"input": "3\n    acornistnt\n    peanutbomb\n    constraint", "output": "1\n    \n\ns_1 = `acornistnt` is an anagram of s_3 = `constraint`. There are no other\npairs i, j such that s_i is an anagram of s_j, so the answer is 1.\n\n* * *"}, {"input": "2\n    oneplustwo\n    ninemodsix", "output": "0\n    \n\nIf there is no pair i, j such that s_i is an anagram of s_j, print 0.\n\n* * *"}, {"input": "5\n    abaaaaaaaa\n    oneplustwo\n    aaaaaaaaba\n    twoplusone\n    aaaabaaaaa", "output": "4\n    \n\nNote that the answer may not fit into a 32-bit integer type, though we cannot\nput such a case here."}]