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 either::*;
use proconio::input;
use proconio::marker::{Chars, Usize1};

fn main() {
    input! {
        n: usize,
        mut s: Chars,
        q: usize,
        //queries: [(usize, char, char); q],
    }
    let mut queries: Vec<(usize, usize, Either<usize, char>)> = vec![];
    for _ in 0..q {
        input! {
            k: usize,
        }
        if k == 1 {
            input! {
                i: Usize1,
                c: char,
            }
            queries.push((k, i, Right(c)))
        } else if k == 2 {
            input! {
                l: Usize1,
                r: Usize1,
            }
            queries.push((k, l, Left(r)));
        }
    }

    let mut bits: Vec<BIT<isize>> = vec![BIT::new(n); 26];
    for i in 0u8..26 {
        let c = ('a' as u8 + i) as char;
        for (i, ns) in s.iter().enumerate() {
            if *ns == c {
                let index = (*ns as u8 - 'a' as u8) as usize;
                bits[index].add(i, 1);
            }
        }
    }

    for &query in &queries {
        if query.0 == 1 {
            update_bits(&s, &mut bits, query.1, query.2.right().unwrap());
        } else if query.0 == 2 {
            let ans = calc(&bits, query.1, query.2.left().unwrap());
            println!("{}", ans);
        }
    }
}

fn update_bits(s: &Vec<char>, bits: &mut Vec<BIT<isize>>, from: usize, to: char) {
    let c = (s[from] as u8 - 'a' as u8) as usize;
    bits[c].add(from, -1);
    let index = (to as u8 - 'a' as u8) as usize;
    bits[index].add(from, 1);
}

fn calc(bits: &Vec<BIT<isize>>, l: usize, r: usize) -> usize {
    //println!("l {}, r {}", l, r);
    let mut count = 0;
    for bit in bits {
        if bit.sum(l as isize, (r + 1) as isize) != 0 {
            count += 1;
        }
    }
    count
}

use num::Num;

#[derive(Clone)]
struct BIT<T> {
    pub n: usize,
    pub d: Vec<T>,
}

impl<T> BIT<T>
where
    T: Num + Copy + Clone,
{
    fn new(n: usize) -> Self {
        BIT {
            n: n,
            d: vec![T::zero(); n + 1],
        }
    }

    fn add(&mut self, mut i: usize, x: T) {
        i += 1;
        while i <= self.n {
            self.d[i] = self.d[i] + x;
            i += i & i.wrapping_neg();
        }
    }

    fn sum_one(&self, mut i: isize) -> T {
        if i < 0 {
            return T::zero();
        }
        let mut x = T::zero();
        i += 1;
        while i != 0 {
            x = x + self.d[i as usize];
            i -= i & i.wrapping_neg();
        }
        x
    }

    fn sum(&self, l: isize, r: isize) -> T {
        self.sum_one(r - 1) - self.sum_one(l - 1)
    }
}