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: Time Limit Exceeded)

import java.util.*;
public class Main {
	private static int ans[];
	public static void main(String[] args){
    	Scanner sc = new Scanner(System.in);
    	int n = sc.nextInt();
    	String s = sc.next();
    	int q = sc.nextInt();

    	for (int i=0;i<q;i++) {
    		int a = sc.nextInt();
    		if (a==1) {
    			int b = sc.nextInt();
    			String c = sc.next();
    			s = s.substring(0,b-1)+c+s.substring(b);
    		}else{
    			int d = sc.nextInt();
    			int e = sc.nextInt();
              	String temp=s.substring(d-1,e);
                System.out.println(count(temp));
    		}
    	}
    }
    	private static int count(String s) {
    		String wk = s;
    		int ret = 0;
    			int l = wk.length();
    			if (l-wk.replace("a","").length()>0) ret++;
    			if (l-wk.replace("b","").length()>0) ret++;
    			if (l-wk.replace("c","").length()>0) ret++;
    			if (l-wk.replace("d","").length()>0) ret++;
    			if (l-wk.replace("e","").length()>0) ret++;
    			if (l-wk.replace("f","").length()>0) ret++;
    			if (l-wk.replace("g","").length()>0) ret++;
    			if (l-wk.replace("h","").length()>0) ret++;
    			if (l-wk.replace("i","").length()>0) ret++;
    			if (l-wk.replace("j","").length()>0) ret++;
    			if (l-wk.replace("k","").length()>0) ret++;
    			if (l-wk.replace("l","").length()>0) ret++;
    			if (l-wk.replace("m","").length()>0) ret++;
    			if (l-wk.replace("n","").length()>0) ret++;
    			if (l-wk.replace("o","").length()>0) ret++;
    			if (l-wk.replace("p","").length()>0) ret++;
    			if (l-wk.replace("q","").length()>0) ret++;
    			if (l-wk.replace("r","").length()>0) ret++;
    			if (l-wk.replace("s","").length()>0) ret++;
    			if (l-wk.replace("t","").length()>0) ret++;
    			if (l-wk.replace("u","").length()>0) ret++;
    			if (l-wk.replace("v","").length()>0) ret++;
    			if (l-wk.replace("w","").length()>0) ret++;
    			if (l-wk.replace("x","").length()>0) ret++;
    			if (l-wk.replace("y","").length()>0) ret++;
    			if (l-wk.replace("z","").length()>0) ret++;
    		return ret;
    	}
}