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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int seg[26][1100006];
int m;
void set(int c, int x, int v)
{
	x += m - 1;
	seg[c][x] = v;
	while (x > 0)
	{
		x = (x - 1) / 2;
		seg[c][x] = seg[c][2 * x + 1] + seg[c][2 * x + 2];
	}
	return;
}
int get(int c, int l, int r)
{
	int res = 0;
	l += m - 1;
	r += m - 1;
	while (l < r)
	{
		if (l % 2 == 0)
		{
			res += seg[c][l];
			l++;
		}
		if (r % 2 > 0)
		{
			res += seg[c][r];
			r--;
		}
		l = (l - 1) / 2;
		r = (r - 1) / 2;
	}
	if (l == r)
		res += seg[c][l];
	return res;
}
int main()
{
	int n;
	scanf("%d", &n);
	char s[500005];
	scanf("%s", s);
	int q;
	scanf("%d", &q);
	int i, j;
	m = 1;
	while (m < n)
		m *= 2;
	for (j = 0; j < 26; j++)
	{
		for (i = 0; i < 2 * m - 1; i++)
			seg[j][i] = 0;
		for (i = 0; i < n; i++)
			if (s[i] - 'a' == j)
				seg[j][i + m - 1] = 1;
		for (i = m - 2; i >= 0; i--)
			seg[j][i] = seg[j][2 * i + 1] + seg[j][2 * i + 2];
	}
	int ans;
	int qq, q1, q2;
	char cq;
	for (i = 0; i < q; i++)
	{
		scanf("%d", &qq);
		if (qq == 1)
		{
			scanf("%d %c", &q1, &cq);
			q1--;
			set(s[q1] - 'a', q1, 0);
			set(cq - 'a', q1, 1);
		}
		else
		{
			scanf("%d %d", &q1, &q2);
			q1--;
			q2--;
			ans = 0;
			for (j = 0; j < 26; j++)
				if (get(j, q1, q2) > 0)
					ans++;
			printf("%d\n", ans);
		}
	}
	return 0;
}