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>
#define ll long long int
#define MOD 1000000007
#define INF 1e18
#define PI 3.14159265358979

using namespace std;

ll gcd(ll a, ll b);
ll power(ll n, ll k);
ll inv(ll n);
ll genfact(ll n, ll k);
ll fact(ll n);
ll nck(ll n, ll k);
vector<vector<ll>> seg;
ll N;
string s;

void update(int b, char c){
  seg[b + N - 1][s[b] - 'a']--;
  seg[b + N - 1][c - 'a']++;
  s[b] = c;
  b = b + N - 1;
  while (b > 0){
    b = (b - 1) / 2;
    for (int i = 0; i < 26; i++){
      seg[b][i] = seg[b * 2 + 1][i] | seg[b * 2 + 2][i];
    }
  }
}

vector<ll> query(int a, int b, int l, int r, int k){
  if (r <= a || b <= l) return vector<ll>(26, 0);
  if (a <= l && r <= b){
    return seg[k]; // ノード k の担当範囲がクエリ区間 [a, b) に完全に含まれる
  }
  else {
    // 一部だけ範囲が被る場合は左右の子に委託する
    vector<ll> c1 = query(a, b, l, (l + r) / 2, 2 * k + 1); // 左の子に値を聞く
    vector<ll> c2 = query(a, b, (l + r) / 2, r, 2 * k + 2); // 右の子に値を聞く
    vector<ll> c3(26, 0);
    for (int i = 0; i < 26; i++){
      c3[i] = c1[i] | c2[i];
    }
    return c3; // 左右の子の値の min を取る
  }
}

int main(void){
  int n;
   cin >> n;
   cin >> s;

  N = 1;
  while (N < n)
    N *= 2;


  seg = vector<vector<ll>>(2*N - 1, vector<ll>(26, 0));

  for (int i = 0; i < n; i++){
    seg[i + N - 1][s[i] - 'a']++;
    update(i, s[i]);
  }


  int q; cin >> q;
  int a, b, l, r;
  char c;
  for (int i = 0; i < q; i++){
    cin >> a;
    if (a == 1){
      cin >> b >> c;
      b--;
      update(b, c);
    }
    else {
      cin >> l >> r;
      l--; r--;
      vector<ll> d = query(l, r + 1, 0, N, 0);
      int ret = 0;
      for (int j = 0; j < 26; j++){
        if (d[j] > 0)
          ret++;
      }
      cout << ret << endl;
    }
  }

  return 0;
}

ll gcd(ll a, ll b){
  if (a < b)
    return gcd(b, a);
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}

ll power(ll n, ll k){
  ll res = 1;
  while(k > 0){
    if (k & 1) 
      res = res * n % MOD;
    n = n * n % MOD;
    k >>= 1;
  }
  return res;
}

ll genfact(ll n, ll k){

  ll res = 1;
  while (k >= 0){
    res = (res * (n - k)) % MOD;
    k--;
  }

  return res;
}

ll inv(ll n){
  return power(n, MOD - 2);
}

ll fact(ll n){
  return genfact(n, n - 1);
}

ll nck(ll n, ll k){
  if (k == 0)
    return 1;
  if (n <= 0)
    return 0;

  return ((genfact(n, k) % MOD) * inv(fact(k))) % MOD;
}