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: Accepted)

N = gets.to_i
class S
	def initialize(s)
		@l = 2 ** (N - 1).bit_length
		@heap = Array.new(@l * 2){0}
		a = "a".ord
		N.times{|i|
			@heap[@l + i] = 1 << (s[i].ord - a)
		}
		(@l - 1).downto(1){|i|
			@heap[i] = @heap[i * 2] | @heap[i * 2 + 1]
		}
		#warn @heap.map{|i|"%x" % i}.inspect
	end
	def replace(i, c)
		#warn "replace(#{i}, #{c})"
		l = @l + i
		c = 1 << (c.ord - "a".ord)
		return if @heap[l] == c
		@heap[l] = c
		l >>= 1
		while l > 0
			tmp = @heap[l]
			@heap[l] = @heap[l * 2] | @heap[l * 2 + 1]
			break if tmp == @heap[l]
			l >>= 1
		end
		#warn @heap.inspect
	end
	def count_r(l, r, n = 1, min = 0, max = @l - 1)
		#warn "count_r(#{l}, #{r}, #{n}, #{min}, #{max})"
		if l <= min and r >= max
			return @heap[n]
		end
		result = 0
		c = (min + max+ 1) / 2
		if l < c
			result |= count_r(l, r, n * 2, min, c - 1)
		end
		if r >= c
			result |= count_r(l, r, n * 2 + 1, c, max)
		end
		result
	end
	def count(l, r)
		b = count_r(l, r)
		c = 0
		while b > 0
			c += 1 if b & 1 == 1
			b >>= 1
		end
		c
	end
end
$s = S.new(gets)
Q = gets.to_i
Q.times{
	q = gets.split
	case q[0]
	when "1"
		$s.replace(q[1].to_i - 1, q[2])
	when "2"
		puts $s.count(q[1].to_i - 1, q[2].to_i - 1)
	end
}