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())

ss = list(input())
s = []
for i in range(n):
    s.append(1<<(ord(ss[i])-97))

#####segfunc######
def segfunc(x,y):
    #print(x|y)
    return x | y

def init(init_val):
    #set_val
    for i in range(n):
        seg[i+num-1]=init_val[i]
    #built
    for i in range(num-2,-1,-1) :
        seg[i]=segfunc(seg[2*i+1],seg[2*i+2]) 
    
def update(k,x):
    k += num-1
    seg[k] = x
    while k:
        k = (k-1)//2
        seg[k] = segfunc(seg[k*2+1],seg[k*2+2])
    
def query(p,q,k,l,r):
    if r<=p or q<=l:
        return ide_ele
    if p<=l and r<=q:
        return seg[k]
    else:
        vl = query(p,q,k*2+1,l,(r+l)//2)
        vr = query(p,q,k*2+2,(l+r)//2,r)
        return segfunc(vl,vr)

#####単位元######
ide_ele = 0

#num:n以上の最小の2のべき乗
num =2**(n-1).bit_length()
seg=[ide_ele]*2*num
init(s)


def popcnt(n):
    c = (n & 0x5555555555555555) + ((n>>1) & 0x5555555555555555)
    c = (c & 0x3333333333333333) + ((c>>2) & 0x3333333333333333)
    c = (c & 0x0f0f0f0f0f0f0f0f) + ((c>>4) & 0x0f0f0f0f0f0f0f0f)
    c = (c & 0x00ff00ff00ff00ff) + ((c>>8) & 0x00ff00ff00ff00ff)
    c = (c & 0x0000ffff0000ffff) + ((c>>16) & 0x0000ffff0000ffff)
    c = (c & 0x00000000ffffffff) + ((c>>32) & 0x00000000ffffffff)
    return c

q = int(input())

for i in range(q):
    qq,xx,yy = map(str,input().split())
    if qq == '1':
        xx = int(xx)
        update(xx-1,1<<(ord(yy)-97))
    else:
        xx,yy = int(xx),int(yy)
        res = query(xx-1,yy,0,0,num)
        print(popcnt(res))