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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#ifdef LOCAL
#define debug(var) cerr << #var << " = " << var;
#else
#define debug(var)
#endif

void init() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}

const int N = 5e5+23;
const int sqN = 800;

int n, q;
string s;
int a[710][30];


void solve() {
  cin >> n >> s >> q;
  for (int i = 0; i < n; ++i) {
    ++a[i/sqN][s[i]-'a'];
  }
  while (q--) {
    int t; cin >> t;
    if (t == 1) {
      int pos;
      char c;
      cin >> pos >> c; --pos;
      c -= 'a';
      --a[pos/sqN][s[pos]-'a'];
      ++a[pos/sqN][c];
      s[pos] = c+'a';
    } else {
      int l, r; cin >> l >> r;
      --l; --r;
      int ans = 0;
      int h[30] = {0};
      for (int i = l; i <= r;) {
	if (i % sqN == 0 && i+sqN-1 <= r) {
	  for (int i = 0; i < 30; ++i) {
	    h[i] += a[i/sqN][i];
	  }
	  i += sqN;
	} else {
	  ++h[s[i]-'a'];
	  ++i;
	}
      }
      for (int i = 0; i < 30; ++i) if (h[i]) ++ans;
      printf("%d\n", ans);
    }
  }
}

int main() {
  init();
  int t = 1; //scanf("%d", &t);
  while (t--) {
    solve();
  }
  return 0;
}