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.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);};
	public static void main(String[] args) {
	int N=sc.nextInt();
	String S =sc.next();
	int Q=sc.nextInt();
	
	for(int i=0;i<Q;i++) {
		int type=sc.nextInt();
		int a=sc.nextInt();
		if(type==1) {
			String b=sc.next();
			StringBuilder sb = new StringBuilder(S);
			sb.setCharAt(a-1,b.charAt(0));
			S=sb.toString();
		}else if(type==2) {
			int b=sc.nextInt();
			int cnt=0;
			//a~bまでで何文字種類があるか
			//charat(a-1)~charat(b-1)
			int mojiretu[]=new int[N+100];
			for(int g=a-1;g<=b-1;g++) {
				if(mojiretu[S.charAt(g)]==0) {
					mojiretu[S.charAt(g)]++;
					cnt++;
				}
				
			}
			p(cnt);
			
		}
	}
	
	}
}