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<iostream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;

class Fenwick {
public:
  	vector<vector<int>> f;
  	string str;
  
    Fenwick(int n, char s[]) {
      	vector<vector<int>> v(n + 1, vector<int>(26));
      	swap(f, v);
      	//f.resize(n + 1);
      	str = string(s);
      	//for (int i = 0; i <= n; ++i) f[i].resize(26);
      	for (int i = 0; i < n; ++i)
          for (int j = i + 1; j <= n; j += j & (-j))
            f[j][s[i] - 'a']++;
    }
    
    void update(int n, char c) {
      	if (c == str[n-1]) return;
      	for (int i = n; i < f.size(); i += i & (-i)) {
          f[i][c - 'a']++;
          f[i][str[n-1] - 'a']--;
        }
      	str[n-1] = c;
    }
  
    void sum(int n, int arr[]) {
      	for (int i = n; i > 0; i -= i & (-i))
          for (int j = 0; j < 26; ++j)
            arr[j] += f[i][j];
    }
  
    int query(int l, int r) {
      int qa[26] = {}, qb[26] = {}, s(0);
      sum(l - 1, qa);
      sum(r, qb);
      for (int i = 0; i < 26; ++i) 
        if (qa[i] != qb[i]) s++;
      return s;
    }
};

int main() {
  int n, q, t, x, y;
  cin >> n;
  char s[n], c;
  for (int i = 0; i < n; ++i) cin >> s[i];
  Fenwick f(n, s);
  cin >> q;
  while (q--) {
    cin >> t >> x;
    if (t == 1) {
      cin >> c;
      f.update(x, c);
    } else {
      cin >> y;
      cout << f.query(x, y) << '\n';
    }
  }
  return 0;
}