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 ST
	def initialize a
		@l = (1<<(a.size-1).bit_length)-1
		@h = [nil]*@l + a
		@h << a[-1] if a.size&1==1
		(@l-1).downto(0){|i|
			j = 2*i+1
			next unless @h[j]
			@h[i] = @h[j]|(@h[j+1]||0)
		}
	end

	def [] l,r=l
		x,l,r = 0,l+@l,r+@l
		x,l,r = x|@h[l]|@h[r],l/2,r/2-1 while l<=r
		return x
	end

	def []= i,x
		i += @l
		@h[i] = x
		while 0<i
			i = (i-1)/2
			j = 2*i+1
			@h[i] = @h[j]|(@h[j+1]||0)
		end
	end
end

gets
bs = ST.new [0]+gets.chomp.bytes.map{|c| 1<<(c-?a.ord) }
gets.to_i.times{
	c,d,e = gets.split
	case c
	when ?1
		bs[d.to_i] = 1<<e.ord-?a.ord

	when ?2
		p 26.times.inject([0,bs[d.to_i,e.to_i]]){|(n,b),| next n+(b&1),b>>1 }[0]
	end
}