Problem:
Score : 500 points

Problem 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).

Constraints

N, Q, i_q, l_q, and r_q are integers.

S is a string consisting of lowercase English letters.

c_q is a lowercase English letter.

1 \leq N \leq 500000

1 \leq Q \leq 20000

|S| = N

1 \leq i_q \leq N

1 \leq l_q \leq r_q \leq N

There is at least one query of type 2 in each testcase.

Input

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

Output

For each query of type 2, print a line containing the answer.

Sample Input 1

7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7

Sample Output 1

3
1
5

In the first query, cdbb contains three kinds of letters: b , c , and d, so we print 3.

In the second query, S is modified to abcdzbd.

In the third query, a contains one kind of letter: a, so we print 1.

In the fourth query, S is modified to abcazbd.

In the fifth query, S does not change and is still abcazbd.

In the sixth query, abcazbd contains five kinds of letters: a, b, c, d, and z, so we print 5.

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

class Bit
  def initialize(n)
    @bit = Array.new(n + 1, 0)
    @n = n
  end

  # 0-origin
  def add(i, x)
    i += 1
    while i <= @n
      @bit[i] += x
      i += i & -i
    end
  end

  # 0-origin
  def sum(i)
    i += 1
    s = 0
    while i > 0
      s += @bit[i]
      i -= i & -i
    end
    s
  end
end

N = gets.to_i
S = gets.chomp.chars.map { |ch| ch.ord - 97 }
bits = Array.new(26) { Bit.new(N) }
S.each.with_index do |o, i|
  bits[o].add(i, 1)
end
Q = gets.to_i
Q.times do
  k, a, b = gets.chomp.split
  a = a.to_i - 1
  k = k.to_i
  if k == 1
    b = b.ord - 97
    next if S[a] == b
    prev = S[a]
    S[a] = b
    bits[prev].add(a, -1)
    bits[b].add(a, 1)
  else
    b = b.to_i - 1
    ans = 0
    26.times do |i|
      ans += 1 if bits[i].sum(b) - (a == 0 ? 0 : bits[i].sum(a - 1)) > 0
    end
    puts ans
  end
end