#include<bits/stdc++.h>
using namespace std;
const int ALPHABET_SIZE = 26;
set<int> pos[ALPHABET_SIZE];
string s;
int n;
int main(int argc, char const *argv[])
{
	cin >> n;
	cin >> s;
	n = s.length();
	for (int i = 0; i < n; ++i)
	{
		pos[s[i] - 'a'].insert(i);
	}
	int q;
	cin >> q;
	while (q --)
	{
		int t;
		cin >> t;
		if (t == 1)
		{
			int l;
			char ch;
			cin >> l >> ch;
			l --;
			pos[s[l] - 'a'].erase(l);
			s[l] = ch;
			pos[s[l] - 'a'].insert(l);
		}
		else
		{
			int l, r, ans = 0;
			cin >> l >> r;
			l --;
			r --;
			for (int i = 0; i < 26; ++i)
			{
				auto it = pos[i].lower_bound(l);
				if (it == pos[i].end() || *it > r)
					continue;
				else
					ans ++;
			}
			cout << ans << '\n';
		}
	}
	return 0;
}
### 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;
const int ALPHABET_SIZE = 26;
set<int> pos[ALPHABET_SIZE];
string s;
int n;
int main(int argc, char const *argv[])
{
	cin >> n;
	cin >> s;
	n = s.length();
	for (int i = 0; i < n; ++i)
	{
		pos[s[i] - 'a'].insert(i);
	}
	int q;
	cin >> q;
	while (q --)
	{
		int t;
		cin >> t;
		if (t == 1)
		{
			int l;
			char ch;
			cin >> l >> ch;
			l --;
			pos[s[l] - 'a'].erase(l);
			s[l] = ch;
			pos[s[l] - 'a'].insert(l);
		}
		else
		{
			int l, r, ans = 0;
			cin >> l >> r;
			l --;
			r --;
			for (int i = 0; i < 26; ++i)
			{
				auto it = pos[i].lower_bound(l);
				if (it == pos[i].end() || *it > r)
					continue;
				else
					ans ++;
			}
			cout << ans << '\n';
		}
	}
	return 0;
}
```