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)

N = gets.to_i
class S
    D = 1000
    def initialize(s)
        @s = s.scan(/./)
        @cache = Array.new(((N - 1) / D) + 1){|i|
            @s[i * D, D].uniq
        }
    end
    def replace(i, c)
        return if @s[i] == c
        @s[i] = c
        @cache[i / D] = @s[i / D * D, D].uniq
        nil
    end
    def count(l, r)
        #warn [l, r].inspect
        result = []
        if l / D == r / D
            result |= @s[l .. r]
        else
            result |= @s[l .. ((l / D * D) + D - 1)]
            result |= @s[(r / D * D) .. r]
            ((l / D + 1) .. (r / D - 1)).each{|i|
                result |= @cache[i]
            }
        end
        #warn result.inspect
        puts result.size
    end
end
$s = S.new(gets.chomp)
#warn s.inspect

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"
        $s.count(q[1].to_i - 1, q[2].to_i - 1)
    end
}