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: Wrong Answer)

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();
		int q=sc.nextInt();
		TreeSet<Integer>[] set=new TreeSet[26];
		for(int i=0; i<26; i++){
			set[i]=new TreeSet<>();
		}
		for(int i=0; i<n; i++){
			set[s[i]-'a'].add(i);
		}
		for(int i=0; i<q; i++){
			int a=sc.nextInt();
			if(a==1){
				int b=sc.nextInt()-1;
				char c=sc.next().charAt(0);
				set[s[b]-'a'].remove(b);
				set[c-'a'].add(i);
				s[b]=c;
			}else{
				int b=sc.nextInt()-1;
				int c=sc.nextInt()-1;
				int ans=0;
				for(int j=0; j<26; j++){
					 Integer ch=set[j].ceiling(b);
         			 if(ch!=null&&ch<=c){
            			ans++;
          			}
				}
				System.out.println(ans);
			}
		}
	}
}