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>
#include <type_traits>
using namespace std;

using ll = long long;

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rrep(i, n) for (int i = 1; i <= (n); ++i)
#define drep(i, n) for (int i = (n)-1; i >= 0; --i)
#define ddrep(i, n) for (int i = n; i > 0; --i)
#define srep(i, s, t) for (int i = s; i < t; ++i)
#define ssrep(i, s, t) for (int i = s; i <= t; ++i)

#define rng(a) a.begin(), a.end()
#define rrng(a) a.rbegin(), a.rend()
#define pb push_back
#define eb emplace_back
#define fi first
#define se second

#define chmax(x, y) (x = max(x, y))
#define chmin(x, y) (x = min(x, y))

using pi = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using ld = long double;

template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}

template <class T> ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  rep(i, (int)v.size()) {
    if (i)
      os << ",";
    os << v[i];
  }
  os << "}";
  return os;
}

template <typename T, size_t S> void printArray(const T (&array)[S]) {
  for (auto val : array)
    std::cout << val << ", ";
  std::cout << "\n";
}

const int mod = 1e9 + 7;
const int inf = 1e9 + 5;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << std::setprecision(10);

  int n; std::cin >> n;
  string s; std::cin >> s;
  int len = s.size();
  vector<set<int>> pos(26); rep(i, len) pos[s[i]-'a'].insert(i);

  int q; std::cin >> q;
  rep(i, q) {
    int qq; std::cin >> qq;
    if (qq == 1) {
      int iq;
      char cq;
      std::cin >> iq >> cq; iq--;
      if (s[iq] == cq) continue;

      char t = s[iq];
      pos[t-'a'].erase(iq);
      pos[cq-'a'].insert(iq);
      s[iq] = cq;
    } else {
      int l, r; std::cin >> l >> r;
      --l, --r;

      int ans = 0;
      rep(j, 26) {
        auto it = pos[j].lower_bound(l);
        if (it != pos[j].end() && *it <= r)
          ans++;
      }
      std::cout << ans << "\n";
    }
  }

}