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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.PrintWriter;
public class Main{
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		char[] s=sc.next().toCharArray();
		BIT[] count=new BIT[26];
		for(int j=0; j<26; j++){
			count[j]=new BIT(1+s.length);
		for(int i=1; i<=s.length; i++){
			if((s[i-1]-'a')==j){
				count[j].add(i,1);
			}
		}
		}
		int q=sc.nextInt();
		for(int i=0; i<q; i++){
			int a=sc.nextInt();
			int b=sc.nextInt();
			if(a==1){
				char c=sc.next().charAt(0);
				count[c-'a'].add(b,1);
				count[s[b-1]-'a'].add(b,-1);
				s[b-1]=c;
			}else{
				int c=sc.nextInt();
				int ans=0;
				for(int j=0; j<26; j++){
					if(count[j].get(c)-count[j].get(b-1)>0){
						ans++;
					}
				}
				System.out.println(ans);
			}
		}
	}
	static class BIT{
		int n;
		int[] data;
		BIT(int n){
			this.n=n;
			data = new int[n+1];
		}
		void add(int i,int v){
			for(;i<=n;i+=i&(-i))data[i]+=v;
		}
		int get(int i){
			int res = 0;
			for(;i>0;i-=i&(-i))res+=data[i];
			return res;
		}
	}
}