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: Time Limit Exceeded)

// atcoder/abc157/E/main.cpp
// author: @___Johniel
// github: https://github.com/johniel/

#include <bits/stdc++.h>

#define each(i, c) for (auto& i : c)
#define unless(cond) if (!(cond))

using namespace std;

template<typename P, typename Q> ostream& operator << (ostream& os, pair<P, Q> p) { os << "(" << p.first << "," << p.second << ")"; return os; }
template<typename P, typename Q> istream& operator >> (istream& is, pair<P, Q>& p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator << (ostream& os, vector<T> v) { os << "("; for (auto& i: v) os << i << ","; os << ")"; return os; }
template<typename T> istream& operator >> (istream& is, vector<T>& v) { for (auto& i: v) is >> i; return is; }
template<typename T> ostream& operator << (ostream& os, set<T> s) { os << "#{"; for (auto& i: s) os << i << ","; os << "}"; return os; }
template<typename K, typename V> ostream& operator << (ostream& os, map<K, V> m) { os << "{"; for (auto& i: m) os << i << ","; os << "}"; return os; }

template<typename T> inline T setmax(T& a, T b) { return a = std::max(a, b); }
template<typename T> inline T setmin(T& a, T b) { return a = std::min(a, b); }

using lli = long long int;
using ull = unsigned long long;
using point = complex<double>;
using str = string;
template<typename T> using vec = vector<T>;

constexpr lli mod = 1e9 + 7;

template<typename T>
struct SegTree {
  int n;
  vector<T> dat;
  using F = function<T(T, T)>;
  F fn;
  T e;
  SegTree(int n_, T e_, F fn_) {
    e = e_;
    fn = fn_;
    n = 1;
    while (n < n_) n *= 2;
    dat.resize(2 * n - 1, e);
  }

  void update(size_t k, T a) {
    k += n - 1;
    dat[k] = a;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = fn(dat[k * 2 + 1], dat[k * 2 + 2]);
    }
    return ;
  }

  T operator [] (size_t idx) const {
    return dat[idx + n - 1];
  }

  T query(size_t a, size_t b) {
    return query(a, b, 0, 0, n);
  }

  T query(size_t a, size_t b, size_t k, size_t l, size_t r) {
    if (r <= a || b <= l) return e;
    if (a <= l && r <= b) return dat.at(k);

    T vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
    T vr = query(a, b, k * 2 + 2, (l + r) / 2, r);

    return fn(vl,vr);
  }
};
// SegTree<int> rmq(a.size(), -(1 << 29), [] (int a, int b) { return max(a, b); });
// SegTree<int> rmq(a.size(), +(1 << 29), [] (int a, int b) { return min(a, b); });

int main(int argc, char *argv[])
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  int n, q;
  str s;
  while (cin >> n >> s >> q) {
    const int N = 500000 + 5;
    SegTree<set<char>> freq(N, set<char>(), [] (set<char> a, set<char> b) {
      set<char> c;
      c.insert(a.begin(), a.end());
      c.insert(b.begin(), b.end());
      // cout << c << ' ' << a << ' ' << b << endl;
      return c;
    });

    for (int i = 0; i < s.size(); ++i) {
      set<char> t;
      t.insert(s[i]);
      freq.update(i, t);
    }

    for (int _ = 0; _ < q; ++_) {
      int t;
      cin >> t;
      if (t == 1) {
        int i;
        char c;
        cin >> i >> c;
        --i;
        set<char> m;
        m.insert(c);
        freq.update(i, m);
      }
      if (t == 2) {
        int begin, end;
        cin >> begin >> end;
        --begin;
        cout << freq.query(begin, end).size() << endl;
      }
    }
  }

  return 0;
}