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.Arrays;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		Main mainObj = new Main();
		mainObj.solve();
	}
	
	public void solve() {
		Scanner sc = new Scanner(System.in);
		int n = Integer.parseInt(sc.nextLine());
		String str = sc.nextLine();
		int q = Integer.parseInt(sc.nextLine());
		
		SegTree segTree = new SegTree(str);
		
		for(int i = 0; i < q; i++) {
			String[] queryArr = sc.nextLine().split(" ");
			int queryNo = Integer.parseInt(queryArr[0]);
			if(queryNo == 1) {
				int pos = Integer.parseInt(queryArr[1]) - 1;
				char val = queryArr[2].charAt(0);
				segTree.update(val, pos);
			}else if(queryNo == 2){
				
				int left = Integer.parseInt(queryArr[1]) - 1;
				int right = Integer.parseInt(queryArr[2]);
				
				System.out.println(testBit(segTree.query(left, right, 0, 0, segTree.leafSize)));
			}
		}
		sc.close();
	}
	
	
	class SegTree{
		int nodeNum;
		int leafSize;
		int[] data;
		
		public SegTree(String str) {
			
			nodeNum = str.length();
			leafSize = 1;
			while(leafSize < nodeNum) {
				leafSize *= 2; 
			}
			
			data = new int[leafSize * 2 - 1];
			
			for(int i = 0; i < str.length(); i++) {
				update(str.charAt(i), i);
			}
		}
		
		public void update(char val, int pos) {
			
			int pointer = leafSize+pos-1;
			int beforeVal = data[pointer];
			int charInt = val - 97;
			int nowVal = (1 << charInt);
			
			if(beforeVal == nowVal) {
				return;
			}
			
			data[pointer] = nowVal;
			
			while(pointer > 0) {
				pointer = (pointer-1)/2;
				data[pointer] = (data[pointer] & ~beforeVal);
				data[pointer] = (data[pointer] | nowVal);
			}
		}
		
		public int query(int a, int b, int k , int l, int r) {
			if(r <= a || b <= l) {
				return 0;
			}
			if(a <= l && r <= b) {
				return data[k];
			}
			
			int m = (l+r)/2;
			return (query(a,b, k*2+1, l,m) | query(a, b, k*2+2, m, r));
		}
	}
	
	public int testBit(int bit) {
		int ans = 0;
		for(int i = 0; i < 26; i++) {
			if(((bit>>i) & 1 ) == 1) {
				ans++;
			}
		}
		return ans;
	}

}