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
input = sys.stdin.readline
N = int(input())
S = input().rstrip()
Q = int(input())
qs = [input().split() for i in range(Q)]

def ctoi(c):
    return ord(c) - ord('a')

class BinaryIndexedTree:
    def __init__(self,size):
        self.N = size
        self.bit = [0]*(size+1)
    def add(self,x,w):
        while x <= self.N:
            self.bit[x] += w
            x += (x & -x)
    def sum(self,x):
        ret = 0
        while x > 0:
            ret += self.bit[x]
            x -= (x & -x)
        return ret
bits = [BinaryIndexedTree(N) for _ in range(26)]

for i,c in enumerate(S):
    bits[ctoi(c)].add(i+1,1)

s = list(S)
ans = []
for a,b,c in qs:
    if a=='1':
        x = int(b)
        bits[ctoi(c)].add(x,1)
        bits[ctoi(s[x-1])].add(x,-1)
        s[x-1] = c
    else:
        tmp = 0
        for i in range(26):
            if bits[i].sum(int(c)) - bits[i].sum(int(b)-1) > 0:
                tmp += 1
        ans.append(tmp)
print(*ans, sep='\n')