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 Python (3.4.3) (judge verdict: Accepted)

from bisect import bisect_left, bisect_right, insort_left
import sys
readlines = sys.stdin.readline
def main():

	n = int(input())
	s = list(input())
	q = int(input())

	d = {}
	flag = {}
	for i in list('abcdefghijklmnopqrstuvwxyz'):
		d.setdefault(i, []).append(-1)
		flag.setdefault(i, []).append(-1)
	for i in range(n):
		d.setdefault(s[i], []).append(i)
	for i in range(q):
		q1,q2,q3 = map(str,input().split())
		if q1 == '1':
			q2 = int(q2) - 1
			if s[q2] != q3:
				insort_left(flag[s[q2]],q2)
				insort_left(d[q3],q2)
				s[q2] = q3
		else:
			ans = 0
			q2 = int(q2) - 1
			q3 = int(q3) - 1
			if q2 == q3:
				print(1)
				continue
			for string,l in d.items():
				res = 0
				if d[string] != [-1]:
					left = bisect_left(l,q2)
					right = bisect_right(l,q3)
				else:
					left = 0
					right = 0
				if string in flag:
					left2 = bisect_left(flag[string],q2)
					right2 = bisect_right(flag[string],q3)
				else:
					left2 = 0
					right2 = 0
				if left != right:
					if right - left > right2 - left2:
						res = 1
				ans += res
				#print(string,l,res)
			print(ans)
if __name__ == '__main__':
	main()