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: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main() {
  int n;
  cin>>n;
  string t;
  cin>>t;
  char s[n];
  for (int i=0; i<n; i++) s[i] = t[i];
  int q;
  cin>>q;
  set<int> m[26];
  for (int i=0; i<n; i++) {
    m[s[i]-'a'].insert(i);
  }
  for (int _=0; _<q; _++) {
    int type;
    cin>>type;
    if (type == 1) {
      int i; char c;
      cin>>i>>c;
      --i;
      char prev = s[i];
      m[prev-'a'].erase(i);
      m[c-'a'].insert(i);
      s[i] = c;
      continue;
    }
    int l, r;
    cin>>l>>r;
    --l; --r;
    int cnt = 0;
    for (int i=0; i<26; i++) {
      char c = 'a' + i;
      auto it = m[c-'a'].lower_bound(l);
      if (it != m[c-'a'].end()) {
        if (*it <= r) {
          ++cnt;
        }
      }
    }
    cout<<cnt<<endl;
  }
}