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.*;

public class Main {

	void run() {

		//
		// 画面入力
		//
		Scanner sc = new Scanner(System.in);
    	int N = sc.nextInt();
    	String S = sc.next();
    	int Q = sc.nextInt();

    	StringBuilder sb = new StringBuilder(S);
    	// char[] charArray = S.toCharArray();

    	for (int i=0; i<Q; i++) {
        	int a=sc.nextInt();
        	int b=sc.nextInt();
        	int c;
        	//char[] d;
        	String d;
        	if(a==1){
        		d=sc.next();
        		sb.replace(b-1, b, d); //index b-1がreplaceされる
        		
            	// d=sc.next().toCharArray();
        		// charArray[b-1] = d[0];
        	}else{
            	c=sc.nextInt();

            	// String subS = sb.substring(b-1, c); //開始箇所, \0付ける箇所 遅い
            	
            	int alCnt = 0;
            	for(char al = 'a'; al <= 'z'; al++){
            		if(S.indexOf(al)!=-1){
            			alCnt++;
            		}
            	}
            	System.out.println(alCnt);
            	
        		/* 遅い
            	Set<Character> uniqueChar = new HashSet<Character>();
        		for(int j=b-1; j <= c-1; j++ ){
        			uniqueChar.add(charArray[j]);
        		}
        		System.out.println(uniqueChar.size());
        		*/
        	}
    	}
    	
    	sc.close();
    	
	}
	
    public static void main(String[] args) {
        new Main().run();
    }
}