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)

NN = 19
X = [0] * ((1<<NN+1)-1)
 
def popcount(x):
    x -= (x >> 1) & 0x55555555
    x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
    x = (x + (x >> 4)) & 0x0f0f0f0f
    x += x >> 8
    x += x >> 16
    return x & 0x3f

def update(a, x):
    i = (1<<NN) - 1 + a
    X[i] = x
    while i:
        i = (i-1) // 2
        X[i] = X[2*i+1] | X[2*i+2]

def rangeor(a, b):
    l = a + (1<<NN)
    r = b + (1<<NN)
    x = 0
    while l < r:
        if l%2:
            x |= X[l-1]
            l += 1
        if r%2:
            r -= 1
            x |= X[r-1]
        l >>= 1
        r >>= 1
    return x

N = int(input())
A = [ord(a) - 97 for a in input()]
for i, a in enumerate(A):
    X[(1<<NN) + i] = 1 << a
for i in range((1<<NN)-1)[::-1]:
    X[i] = X[i*2+1] | X[i*2+2]

Q = int(input())
for _ in range(Q):
    a, b, c = input().rstrip().split()
    if a == "1":
        b = int(b)
        c = ord(c) - 97
        update(b, 1<<c)
    else:
        b, c = int(b), int(c)
        print(popcount(rangeor(b, c+1)))