#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;

int main() {
  int N, Q;
  string s; cin >> N >> s >> Q;
  vector<set<int>> is(26);
  for(int i = 0; i < N; i++) {
    is[s[i]-'a'].insert(i);
  }
  for(int qi = 0; qi < Q; qi++) {
    int type; cin >> type;
    if(type == 1) {
      int i;
      char c;
      cin >> i >> c;
      i--;
      is[s[i]-'a'].erase(i);
      s[i] = c;
      is[s[i]-'a'].insert(i);
    } else {
      int l, r; cin >> l >> r;
      l--;
      int ans = 0;
      for(int i = 0; i < 26; i++) {
        auto it = is[i].lower_bound(l);
        if(it != is[i].end() && *it < r) ans++;
      }
      cout << ans << endl;
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;

int main() {
  int N, Q;
  string s; cin >> N >> s >> Q;
  vector<set<int>> is(26);
  for(int i = 0; i < N; i++) {
    is[s[i]-'a'].insert(i);
  }
  for(int qi = 0; qi < Q; qi++) {
    int type; cin >> type;
    if(type == 1) {
      int i;
      char c;
      cin >> i >> c;
      i--;
      is[s[i]-'a'].erase(i);
      s[i] = c;
      is[s[i]-'a'].insert(i);
    } else {
      int l, r; cin >> l >> r;
      l--;
      int ans = 0;
      for(int i = 0; i < 26; i++) {
        auto it = is[i].lower_bound(l);
        if(it != is[i].end() && *it < r) ans++;
      }
      cout << ans << endl;
    }
  }
  return 0;
}

```