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

int main()
{
  int n;cin>>n;
  string s;cin>>s;
  int i,j;
  int q;cin>>q;
  vector<set<int>> sets(26);
  for(i=0;i<n;i++)
  {
    sets.at(s.at(i)-'a').insert(i);
  }
  for(i=0;i<q;i++)
  {
    int num;cin>>num;
    if(num==1)
    {
      int ind;
      char c;
      cin>>ind>>c;
      for(j=0;j<26;j++)sets.at(j).erase(ind-1);
      sets.at(c-'a').insert(ind-1);
    }
    else if(num==2)
    {
      int l,r;cin>>l>>r;
      int kind=0;
      for(j=0;j<26;j++)
      {
        auto it=sets.at(j).lower_bound(l-1);
        if(it!=sets.at(j).end()&&*it<=r-1)kind++;
      }
      cout<<kind<<endl;
    }
  }
}

### Prompt

Your task is to create a Cpp solution 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()
{
  int n;cin>>n;
  string s;cin>>s;
  int i,j;
  int q;cin>>q;
  vector<set<int>> sets(26);
  for(i=0;i<n;i++)
  {
    sets.at(s.at(i)-'a').insert(i);
  }
  for(i=0;i<q;i++)
  {
    int num;cin>>num;
    if(num==1)
    {
      int ind;
      char c;
      cin>>ind>>c;
      for(j=0;j<26;j++)sets.at(j).erase(ind-1);
      sets.at(c-'a').insert(ind-1);
    }
    else if(num==2)
    {
      int l,r;cin>>l>>r;
      int kind=0;
      for(j=0;j<26;j++)
      {
        auto it=sets.at(j).lower_bound(l-1);
        if(it!=sets.at(j).end()&&*it<=r-1)kind++;
      }
      cout<<kind<<endl;
    }
  }
}

```