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: Time Limit Exceeded)

import java.util.*;
 
 public class Main {
     public static void main(String args[] ) throws Exception {
		 Scanner sc = new Scanner(System.in);
		 String S = sc.nextLine() ;

		 int N = Integer.valueOf( S ) ;
		 S = sc.nextLine() ;
		 char[] arr = new char[N] ;
		 for ( int j=0 ; j<S.length() ; j++ ) 
			 arr[j] = S.charAt(j)  ;
//		System.out.println( S ) ;
//		System.out.println( Arrays.toString( arr ) ) ;

		 S = sc.nextLine() ;
		 int Q = Integer.valueOf( S ) ;
		 for ( int j=0 ; j<Q ; j++ ) {
			 String[] query = sc.nextLine().split(" ") ;
			 int cmd = Integer.valueOf( query[0] ) ;
			 if ( cmd == 1 ) {
				 int  p1  = Integer.valueOf( query[1] ) ;
				 char c1  = query[2].charAt(0) ;
				 arr[p1-1] = c1 ;

			 } else {
				 int p1 = Integer.valueOf( query[1] ) ;
				 int p2 = Integer.valueOf( query[2] ) ;
				 Set<Character> set = new HashSet<>() ; 
				 for ( int x=p1-1 ;x<=p2-1 ; x++ )
					 set.add ( arr[x] ) ;
				 System.out.println( set.size() ) ;
			 }


		 }


	 }
}