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: Accepted)

"""
セグメント木で行きたいところだけど、集合の管理が必要になるため、時間がかかりすぎてしまう。
と思いきや、文字の種類は26種類しかないので、問題ない。
"""
def main():
    import sys
    input = sys.stdin.readline
    from collections import Counter
    class BIT:
        def __init__(self,n):
            self.size = n
            self.tree = [Counter([]) for _ in range(n+1)]
        
        def change(self,i,x,y):
            while i <= self.size:
                if x in self.tree[i]:
                    self.tree[i][x] -= 1
                if y not in self.tree[i]:
                    self.tree[i][y] = 1
                else:
                    self.tree[i][y] += 1
                i += i&-i
    
        def sum(self,i):
            s = Counter([])
            while i > 0:
                s += self.tree[i]
                i -= i&-i
            cnt = 0
            return s


    N = int(input())
    S = list(input())
    Q = int(input())
    bit = BIT(N)
    for i in range(N):
        bit.change(i+1," ",S[i])

    for _ in range(Q):
        q,a,b = input().split()
        if q == "1":
            x = S[int(a)-1]
            S[int(a)-1] = b
            bit.change(int(a),x,b)
        else:
            a = int(a)
            b = int(b)
            count = bit.sum(b) - bit.sum(a-1)
            ans = 0
            for v in count.values():
                if v > 0:
                    ans += 1
            print(ans)
main()