output_description: For each query of type 2, print a line containing the answer.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      S
      Q
      Query_1
      \vdots
      Query_Q
      
  
  Here, Query_i in the 4-th through (Q+3)-th lines is one of the following:
  
      
      
      1 i_q c_q
      
      
      
      2 l_q r_q
attempt: N = int(input())
  S = list(input())
  Q = int(input())
  # print(S) 012345
  do = [input().split() for i in range(Q)]
  answer = []
  check = []
  # print(do)
  for i in range(Q):
      if int(do[i][0]) == 1:
          S[int(do[i][1]) - 1] = do[i][2]
          # print(S)###
      else:
          for j in range(int(do[i][1]) - 1, int(do[i][2])):
              check.append(S[j])
          check.sort()
          # print(check)####
          while True:
              q = 0
              for k in range(len(check) - 2):
                  if check[k] == check[k + 1]:
                      trash = check.pop(k)
                      # print(check)###
                      q = 1
                      break
              if q == 0:
                  break
          # print(check)#####
          answer.append(len(check))
          check = []
  # print(answer)#####
  kotae = ""
  for h in range(len(answer)):
      print(answer[h])
  
problem_description: Statement
  
  You are given a string S of length N consisting of lowercase English letters.
  
  Process Q queries of the following two types:
  
    * Type 1: change the i_q-th character of S to c_q. (Do nothing if the i_q-th character is already c_q.)
    * Type 2: answer the number of different characters occurring in the substring of S between the l_q-th and r_q-th characters (inclusive).
samples: [{"input": "7\n    abcdbbd\n    6\n    2 3 6\n    1 5 z\n    2 1 1\n    1 4 a\n    1 7 d\n    2 1 7", "output": "3\n    1\n    5\n    \n\nIn the first query, `cdbb` contains three kinds of letters: `b` , `c` , and\n`d`, so we print 3.\n\nIn the second query, S is modified to `abcdzbd`.\n\nIn the third query, `a` contains one kind of letter: `a`, so we print 1.\n\nIn the fourth query, S is modified to `abcazbd`.\n\nIn the fifth query, S does not change and is still `abcazbd`.\n\nIn the sixth query, `abcazbd` contains five kinds of letters: `a`, `b`, `c`,\n`d`, and `z`, so we print 5."}]