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;

// Segment-Tree
struct SegmentTree {
  vector<int> dat;
  int n;
  
  // Constructor
  SegmentTree(int _n) {
    init(_n);
  }
  
  // 初期化
  void init(int _n) {
    int x = 1;
    while(x < _n) x *= 2;
    dat.resize(2 * x - 1);
    n = x;
    for(int i = 0; i < 2 * x - 1; i++) dat[i] = 0;
  }
  
  // i番目の値にxを加算
  void update(int i, int x) {
    i += n - 1; // 葉のノード
    dat[i] += x;
    while(i > 0){ // 更新
      i = (i - 1) / 2;
      dat[i] += x;
    }
  }
  
  // [a, b)の和
  // k:ノードの番号 [l, r)がkに対応づいている
  // query(a, b, 0, 0, n)と呼ぶ
  int query(int a, int b, int k, int l, int r) {
    if(r <= a || b <= l) return 0;
    if(a <= l && r <= b) return dat[k];
    else{
      int vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
      int vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
      return vl + vr;
    }
  }
  
  int ret_n() {
    return n;
  }
};

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  int n, q;
  string s;
  cin >> n >> s >> q;
  
  vector<SegmentTree> x(26, SegmentTree(n));
  for (int i = 0; i < n; i++) {
    x.at(s.at(i) - 'a').update(i, 1);
  }
  
  for (int i = 0; i < q; i++) {
    int a;
    cin >> a;
    if (a == 1) {
      int j;
      char c;
      cin >> j >> c;
      j--;
      x.at(s.at(j) - 'a').update(j, -1);
      x.at(c - 'a').update(j, 1);
      s.at(j) = c;
    }
    else {
      int l, r;
      cin >> l >> r;
      int count = 0;
      for (int j = 0; j < 26; j++) {
        if (x.at(j).query(l - 1, r, 0, 0, x.at(j).ret_n()) > 0) count++;
      }
      cout << count << '\n';
    }
  }
}