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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

int main()
{
	int N, Q;
	char S[500001];
	scanf("%d", &N);
	scanf("%s", S);
	scanf("%d", &Q);
	
	int i, j, count[251][26] = {};
	for (i = 0; i < N / 2000; i++) {
		for (j = 0; j < 2000; j++) count[i][S[i*2000+j] - 'a']++;
	}
	for (j = 0; j < N % 2000; j++) count[i][S[i*2000+j] - 'a']++;

	int l, r, q, type, flag[26], ans;
	char c;
	for (q = 0; q < Q; q++) {
		scanf("%d", &type);
		if (type == 1) {
			scanf("%d %c", &i, &c);
			i--;
			count[i/2000][S[i] - 'a']--;
			count[i/2000][c - 'a']++;
			S[i] = c;
		} else {
			scanf("%d %d", &l, &r);
			l--;
			r--;
			for (j = 0; j < 26; j++) flag[j] = 0;
			for (i = (l + 1999) / 2000; i < (r + 1) / 2000; i++) {
				for (j = 0; j < 26; j++) if (count[i][j] > 0) flag[j] = 1;
			}
			for (j = l; j <= r && j % 2000 != 0; j++) flag[S[j] - 'a'] = 1;
			for (j = r; j >= l && (j + 1) % 2000 != 0; j--) flag[S[j] - 'a'] = 1;
			for (j = 0, ans = 0; j < 26; j++) ans += flag[j];
			printf("%d\n", ans);
		}
	}
	
	fflush(stdout);
	return 0;
}