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

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  set <int> alp[26];
  int n;
  cin >> n;
  char c, s[n];
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    alp[s[i] - 'a'].insert(i);
  }
  int q, t, x, y;
  cin >> q;
  while (q--) {
    cin >> t >> x;
    if (t == 1) {
      cin >> c;
      alp[s[x - 1] - 'a'].erase(x - 1);
      alp[c - 'a'].insert(x - 1);
      s[x - 1] = c;
    } else {
      cin >> y;
      int ans = 0;
      for (int i = 0; i < 26; i++) {
        auto it = alp[i].lower_bound(x - 1);
        if (it != alp[i].end() && *it < y) ans++;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
### Prompt

Create a solution in Cpp for the following problem:

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.

Example

Input

7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7


Output

3
1
5

### Response

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

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  set <int> alp[26];
  int n;
  cin >> n;
  char c, s[n];
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    alp[s[i] - 'a'].insert(i);
  }
  int q, t, x, y;
  cin >> q;
  while (q--) {
    cin >> t >> x;
    if (t == 1) {
      cin >> c;
      alp[s[x - 1] - 'a'].erase(x - 1);
      alp[c - 'a'].insert(x - 1);
      s[x - 1] = c;
    } else {
      cin >> y;
      int ans = 0;
      for (int i = 0; i < 26; i++) {
        auto it = alp[i].lower_bound(x - 1);
        if (it != alp[i].end() && *it < y) ans++;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
```