#include<bits/stdc++.h>
using namespace std;
int n,q;
string s;
set<int>st[33];
int main()
{
	cin>>n>>s>>q;
	for(int i=0;i<n;i++)
	{
		st[s[i]-'a'].insert(i);
	}
	while(q--)
	{
		int tp;
		cin>>tp;
		if(tp==1)
		{
			int p;
			char c;
			cin>>p>>c;
			p--;
			st[s[p]-'a'].erase(p);
			st[c-'a'].insert(p);
			s[p]=c;
		}
		else
		{
			int l,r;
			cin>>l>>r;
			l--;
			r--;
			int cnt=0;
			for(int i=0;i<30;i++)
			{
				if(st[i].lower_bound(l)==st[i].end())
				{
					continue;
				}
				if(*st[i].lower_bound(l)<=r)
				{
					cnt++;
				}
			}
			cout<<cnt<<endl;
		}
	}
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 n,q;
string s;
set<int>st[33];
int main()
{
	cin>>n>>s>>q;
	for(int i=0;i<n;i++)
	{
		st[s[i]-'a'].insert(i);
	}
	while(q--)
	{
		int tp;
		cin>>tp;
		if(tp==1)
		{
			int p;
			char c;
			cin>>p>>c;
			p--;
			st[s[p]-'a'].erase(p);
			st[c-'a'].insert(p);
			s[p]=c;
		}
		else
		{
			int l,r;
			cin>>l>>r;
			l--;
			r--;
			int cnt=0;
			for(int i=0;i<30;i++)
			{
				if(st[i].lower_bound(l)==st[i].end())
				{
					continue;
				}
				if(*st[i].lower_bound(l)<=r)
				{
					cnt++;
				}
			}
			cout<<cnt<<endl;
		}
	}
	return 0;
}
```