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 PyPy3 (2.4.0) (judge verdict: Accepted)

import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def main():
    N,S,Q,*q = read().decode().split()
    N = int(N)
    S = list(S)

    c = lambda x: ord(x) - ord('a')
    tree = [SegmentTree(N, lambda x,y:x|y) for _ in range(26)]
    for i, l in enumerate(S):
        tree[c(l)].update(i, 1)

    for t, a, b in zip(*[iter(q)]*3):
        if t == '1':
            a = int(a) - 1
            tree[c(S[a])].update(a, 0)
            tree[c(b)].update(a, 1)
            S[a] = b
        else:
            a, b = int(a)-1, int(b)-1
            ans = 0
            for i in range(26):
                t = tree[i].query(a, b+1)
                ans += 1 if t else 0
            print(ans)

class SegmentTree:
    def __init__(self, n=None, f=max, zero_factory=int, converter=lambda x:x, initial_values=None):
        assert(n or initial_values)
        size = n if n else len(initial_values)
        self.__f, self.__z, self.__c = f, zero_factory, converter
        self.__n = 1 << ((size - 1).bit_length())
        self.__dat = [zero_factory() for _ in range(2 * self.__n)]
        if initial_values:
            zi = self.__get_leaf(0)
            for i, v in enumerate(initial_values): self.__dat[zi + i] = converter(v)
            for i in range(zi-1, -1, -1): self.__dat[i] = f(*self.__get_children(i))

    def update(self, index, value):
        i, v = self.__get_leaf(index), self.__c(value)
        if self.__dat[i] == v: return
        self.__dat[i] = v
        while i:
            i = self.__get_parent(i)
            self.__dat[i] = self.__f(*self.__get_children(i))

    def query(self, from_inclusive, to_exclusive):
        ans = self.__z()
        if to_exclusive <= from_inclusive: return ans
        l, r = self.__get_leaf(from_inclusive), self.__get_leaf(to_exclusive) - 1
        while r - l > 1:
            if self.__is_right(l): ans = self.__f(ans, self.__dat[l])
            if self.__is_left(r): ans, r = self.__f(ans, self.__dat[r]), r-1
            l, r = l//2, self.__get_parent(r)
        ans = self.__f(ans, self.__dat[l])
        if l != r: ans = self.__f(ans, self.__dat[r])
        return ans

    def __get_leaf(self, i): return self.__n + i - 1
    def __get_parent(self, i): return (i-1)//2
    def __get_children(self, i): return (self.__dat[2*i+1], self.__dat[2*i+2])
    def __is_left(self, i): return i&1 == 1
    def __is_right(self, i): return i&1 == 0

if __name__ == '__main__':
    main()