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 (7.3.0) (judge verdict: Wrong Answer)

N = int(input())
S = ["*"]*(N+1)
S[1:] = list(input())
Q = int(input())
q1,q2,q3 = [0]*Q,[0]*Q,[0]*Q
for i in range(Q):
    tmp = input().split()
    q1[i],q2[i] = map(int,tmp[:2])
    if q1[i] == 2:
        q3[i] = int(tmp[2])
    else:
        q3[i] = tmp[2]

class BIT:
    def __init__(self, n, init_list):
        self.num = n + 1
        self.tree = [0] * self.num
        for i, e in enumerate(init_list):
            self.update(i, e)

    #a_kにxを加算
    def update(self, k, x):
        k = k + 1
        while k < self.num:
            self.tree[k] += x
            k += (k & (-k))
        return

    def query1(self, r):
        ret = 0
        while r > 0:
            ret += self.tree[r]
            r -= r & (-r)
        return ret

    #通常のスライスと同じ。lは含み、rは含まない
    def query2(self, l, r):
        return self.query1(r) - self.query1(l)

s_pos = [BIT(N+1,[0]*(N+1)) for _ in range(26)]

for i in range(1,N+1):
    s_pos[ord(S[i]) - ord("a")].update(i,1)

alphabets = [chr(ord("a") + i) for i in range(26)]

for i in range(Q):
    if q1[i] == 1:
        s_pos[ord(S[q2[i]]) - ord("a")].update(q2[i],-1)
        S[q2[i]] = q3[i]
        print(S)
        s_pos[ord(q3[i]) - ord("a")].update(q2[i],1)
    else:
        ans = 0
        for c in alphabets:
            if s_pos[ord(c) - ord("a")].query2(q2[i],q3[i]+1) > 0:
                ans += 1
        print(ans)