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;
using ll = long long;
const ll MOD = 1e9 + 7;
//const ll MOD = 998244353;
//const ll MOD = ;
ll mod(ll A, ll M) {return (A % M + M) % M;}
const ll INF = 1LL << 60;
template<class T> bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template<class T> bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
ll divceil(ll A, ll B) {return (A + (B - 1)) / B;}
ll myctoi(char C) {return C - '0';}
char myitoc(ll N) {return '0' + N;}
#define FINALANS(A) do {cout << (A) << '\n'; exit(0);} while (false)

class BinaryIndexedTree
{
public:
  ll N;
  vector<ll> bit;

  BinaryIndexedTree(ll n)
  {
    N = n;
    bit = vector<ll>(N + 1, 0); // 1 ~ N なので注意
  }

  ll sum(ll i)
  {
    ll s = 0;
    while (i > 0)
    {
      s += bit.at(i);
      i -= i & -i;
    }
    return s;
  }

  void add(ll i, ll x)
  {
    while (i <= N)
    {
      bit.at(i) += x;
      i += i & -i;
    }
  }
};

int main()
{
  using BIT = BinaryIndexedTree;

  ll N;
  cin >> N;
  string S;
  cin >> S;

  vector<BIT> B(26, BIT(N));
  for (ll i = 1; i <= N; i++)
  {
    B.at(S.at(i - 1) - 'a').add(i, 1);
  }

  ll Q;
  cin >> Q;
  for (ll q = 0; q < Q; q++)
  {
    ll qt;
    cin >> qt;
    if (qt == 1)
    {
      ll i;
      char c;
      cin >> i >> c;
      B.at(S.at(i - 1) - 'a').add(i, -1);
      S.at(i - 1) = c;
      B.at(c - 'a').add(i, 1);
    }
    else
    {
      ll l, r;
      cin >> l >> r;

      ll ans = 0;
      for (ll j = 0; j < 26; j++)
      {
        ans += (B.at(j).sum(r) - B.at(j).sum(l - 1) >= 1);
      }

      cout << ans << endl;
    }
  }
}