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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

#!/usr/bin/env python3

from math import ceil

class Segtree:
    def __init__(self, begin, size, s):
        self.begin = begin
        self.size = size
        self.counter = [0 for i in range(26)]  # ttk
        for c in s[begin - 1:begin + size - 1]:
            ci = ord(c) - ord("a")
            self.counter[ci] += 1
        if size == 1:
            self.left = None
            self.right = None
            self.ci = ci
        else:
            sl = ceil(size / 2.0)
            sr = size - sl
            self.left = Segtree(begin, sl, s)
            self.right = Segtree(begin + sl, sr, s)

    def change_ith_to_c(self, i1, ci):
        if self.size == 1:
            assert i1 == self.begin
            prev = self.ci
            self.ci = ci
        else:
            sl = ceil(self.size / 2.0)
            child = self.left if i1 <= self.begin + sl - 1 else self.right
            prev = child.change_ith_to_c(i1, ci)
        self.counter[prev] -= 1
        self.counter[ci] += 1
        return prev

    def qt2(self, l, r):
        assert self.begin <= l
        assert r <= self.begin + self.size - 1
        if self.size == 1:
            return {self.ci}
        sl = ceil(self.size / 2.0)
        center = self.begin + sl - 1  # End of left
        if r <= center:
            return self.left.qt2(l, r)
        elif l > center:
            return self.right.qt2(l, r)
        else:
            return self.left.qt2(l, center) | self.right.qt2(center + 1, r)

    def __repr__(self):
        return "[{} {} {} {}]".format(self.begin, self.begin + self.size - 1,
            repr(self.left), repr(self.right))

def main():
    n = int(input())
    s = input()
    st = Segtree(1, n, s)
    q = int(input())
    for i in range(q):
        t, l, r = input().split()
        if t == "1":
            ci = ord(r) - ord("a")
            st.change_ith_to_c(int(l), ci)
        else:
            print(len(st.qt2(int(l), int(r))))

if __name__ == "__main__":
    main()