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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::input;
fn main() {
    input! {
        n: usize,
        s: String,
        q: usize,
    }
    let w=n.next_power_of_two();
    let mut t=vec![0;w*2];
    for (i,c) in s.bytes().enumerate() {
        t[i+w]=1<<c-b'a';
    }
    for k in (0..w).rev() {
        t[k]=t[k*2]|t[k*2+1];
    }
    for _ in 0..q {
        input! { q: usize }
        if q==1 {
            input! {
                i: usize,
                c: String,
            }
            let mut k=i+w;
            t[k]=1<<c.as_bytes()[0]-b'a';
            while k>0 {
                k/=2;
                t[k]=t[k*2]|t[k*2+1];
            }
        }
        else {
            input! {
                l: usize,
                r: usize,
            }
            println!("{}",f(&t,1,0,w,l-1,r).count_ones());
        }
    }
}
fn f(t: &Vec<u32>, k: usize, a: usize, b: usize, l: usize, r: usize) -> u32 {
    if a==l && b==r {
        t[k]
    }
    else {
        let w=(b-a)/2;
        if l>=a+w {
            f(t,k*2+1,a+w,b,l,r)
        }
        else if r<=a+w {
            f(t,k*2,a,a+w,l,r)
        }
        else {
            f(t,k*2,a,a+w,l,a+w) | f(t,k*2+1,a+w,b,a+w,r)
        }
    }
}