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: Time Limit Exceeded)

from bisect import bisect_left, bisect_right, insort_left

n = int(input())
s = list(input())
q = int(input())
qs = [list(input().split()) for _ in range(q)]

abc = [[-1, n] for _ in range(26)]
for i, e in enumerate(s):
    j = ord(e) - ord("a")
    abc[j].append(i)

for i in range(26):
    abc[i].sort()

for t, e1, e2 in qs:
    if t == "1":
        i = int(e1) - 1
        c = e2

        let = s[i]

        idx = ord(let) - ord("a")
        popi = bisect_left(abc[idx], i)
        abc[idx].pop(popi)

        idx = ord(c) - ord("a")
        insort_left(abc[idx], i)

        s[i] = c

    else:
        l = int(e1) - 1
        r = int(e2) - 1

        cnt = 0
        for i in range(26):
            li = bisect_left(abc[i], l)
            ri = bisect_right(abc[i], r)
            if li != ri:
                cnt += 1

        print(cnt)