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 Java7 (OpenJDK 1.7.0) (judge verdict: Runtime Error)

import java.util.Arrays;
import java.util.Scanner;
 
public class Main {
	private static Scanner sc = new Scanner(System.in);
	static void p(String ans) {System.out.println(ans);};
	static void p(int ans) {System.out.println(ans);};
	static void p(long ans) {System.out.println(ans);};
	static String eikomoji[]={"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};
	public static void main(String[] args) {
	int N=sc.nextInt();
	String S =sc.next();
	int Q=sc.nextInt();
	int type,a,b,cnt=0;
	char v;
	StringBuilder sb = new StringBuilder(S);
	for(int i=0;i<Q;i++) {
		type=sc.nextInt();
		a=sc.nextInt();
		v= sc.next().charAt(0);
		if(type==1) {
			if (sb.charAt(a-1)==v) {
				continue;
			}
			else {
				sb.setCharAt(a-1,v);
			}
		}else{
			b=sc.nextInt();
				for(int f=0;f<=25;f++) {
					if(sb.indexOf(eikomoji[f], a-1)<=b-1&&sb.indexOf(eikomoji[f], a-1)!=-1) {
						cnt++;
					}
				
				
			}
			System.out.println(cnt);
			cnt=0;
			}
	}
	
	}
}