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 (2.7.1) (judge verdict: Accepted)

class SegTree
  def initialize(size, id, fill = id, &op)
    n = 1
    while n < size
      n <<= 1
    end
    @op, @n, @a = op, n, [fill] * (2*n)
  end

  def [](i); @a[i + @n]; end

  def []=(i, x)
    @a[i += @n] = x
    while (i >>= 1) > 0
      @a[i] = @op[@a[i<<1|0], @a[i<<1|1]]
    end
  end

  def fold(l, r = l + 1)
    l += @n
    r += @n
    x = y = @a[0]
    while l < r
      if (l & 1) == 1
        x = @op[x, @a[l]]
        l += 1
      end
      y = @op[@a[r -= 1], y] if (r & 1) == 1
      l >>= 1
      r >>= 1
    end
    @op[x, y]
  end
end

n = gets.to_i
st = SegTree.new(n, 0) { |a, b| a | b }

gets.chomp.each_char.with_index do |c, i|
  x = 1 << (c.ord - 'a'.ord)
  st[i] = x
end

q = gets.to_i
q.times do
  t, x, y = gets.split
  case t.to_i
  when 1
    i, c = x.to_i, y
    x = 1 << (c.ord - 'a'.ord)
    st[i-1] = x
  when 2
    l, r = x.to_i, y.to_i
    x = st.fold(l-1, r)
    cnt = (0...26).count { |b| x[b] == 1 }
    p cnt
  end
end