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

int main() {
  long n,q;
  string s;
  cin >> n >> s >> q;
  vector<set<long>> a(26,set<long>());
  vector<long> ans;
  for(long i=0;i<n;i++) a[s[i]-'a'].insert(i+1);
  for(long i=0;i<q;i++) {
    long q1,q2,q3_2;
    char q3_1;
    cin >> q1 >> q2;
    if(q1==1) {
      cin >> q3_1;
      a[s[q2-1]-'a'].erase(q2);
      s[q2-1]=q3_1;
      a[q3_1-'a'].insert(q2);
    } else {
      cin >> q3_2;
      ans.push_back(0);
      for(long i=0;i<26;i++) {
        auto itr=a[i].lower_bound(q2);
        if(itr!=a[i].end()) if(*itr<=q3_2) ans.back()++;
      }
    }
  }
  for(long i=0;i<ans.size();i++) cout << ans[i] << endl;
}
### Prompt

Please create a solution in cpp to 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() {
  long n,q;
  string s;
  cin >> n >> s >> q;
  vector<set<long>> a(26,set<long>());
  vector<long> ans;
  for(long i=0;i<n;i++) a[s[i]-'a'].insert(i+1);
  for(long i=0;i<q;i++) {
    long q1,q2,q3_2;
    char q3_1;
    cin >> q1 >> q2;
    if(q1==1) {
      cin >> q3_1;
      a[s[q2-1]-'a'].erase(q2);
      s[q2-1]=q3_1;
      a[q3_1-'a'].insert(q2);
    } else {
      cin >> q3_2;
      ans.push_back(0);
      for(long i=0;i<26;i++) {
        auto itr=a[i].lower_bound(q2);
        if(itr!=a[i].end()) if(*itr<=q3_2) ans.back()++;
      }
    }
  }
  for(long i=0;i<ans.size();i++) cout << ans[i] << endl;
}
```