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: Wrong Answer)



class segtree:

    def __init__(self, n,operator,identity):
        nb = bin(n)[2:]
        bc = sum([int(digit) for digit in nb])
        if bc == 1:
            self.num_end_leaves = 2**(len(nb)-1)
        else:
            self.num_end_leaves = 2**(len(nb))

        self.array = [identity for i in range(self.num_end_leaves * 2)]
        self.identity = identity
        self.operator =operator

    def update(self,x,val):
        actual_x = x+self.num_end_leaves
        self.array[actual_x] = val
        while actual_x > 0 :
            actual_x = actual_x//2
            self.array[actual_x] = self.operator(self.array[actual_x*2],self.array[actual_x*2+1])

    def get(self,q_left,q_right,arr_ind=1,leaf_left=0,depth=0):
        width_of_floor = self.num_end_leaves//(2**depth)
        leaf_right = leaf_left+width_of_floor-1

        if leaf_left > q_right or leaf_right < q_left:
            return  self.identity

        elif leaf_left >= q_left and leaf_right <= q_right:
            return self.array[arr_ind]

        else:
            val_l = self.get(q_left,q_right,2*arr_ind,leaf_left,depth+1)
            val_r = self.get(q_left,q_right,2*arr_ind+1,leaf_left+width_of_floor//2,depth+1)
            return self.operator(val_l,val_r)



N=int(input())
S=[i for i in input()]
Q=int(input())

s_tree = segtree(N,(lambda x,y:set(list(x)+list(y))),set([])) 
for i in range(N):
    s_tree.update(i, set([S[i]]))



for _ in range(Q):
    f, a, b =(i for i in input().split())
    if int(f)==1:
        s_tree.update(int(a)-1,set([b]))
    else:
        print(len(s_tree.get(int(a)-1, int(b))))