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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
#define int long long
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> set_t;
set_t ost[26];
int32_t main() {
	ios::sync_with_stdio(0);cin.tie(0);
	int n, q;
	string s;
	cin >> n >> s >> q;
	for(int i=0; i<n; i++) {
		s[i] -= 'a';
		ost[s[i]].insert(i);
	}
	for(int z=0; z<q; z++) {
		int t;
		cin >> t;
		if(t == 1) {
			int i;
			char c;
			cin >> i >> c;
			c -= 'a';
			ost[s[i]].erase(i);
			s[i] = c;
			ost[s[i]].insert(i);
		} else {
			int l, r;
			cin >> l >> r;
			l--, r--;
			int cnt = 0;
			for(int i=0; i<26; i++) {
				cnt += (ost[i].order_of_key(l) != ost[i].order_of_key(r+1));
			}
			cout << cnt << '\n';
		}
	}
}