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

int main() {
	int n; cin >> n;
	string s; cin >> s;
	vector<set<int> > alph(26);
	vector<int> ind(n);

	for (int i = 0; i < n; ++i)
	{
		alph[s[i] - 'a'].insert(i);
		ind[i] = s[i] - 'a';
	}

	int q; cin >> q;
	for (int i = 0; i < q; ++i)
	{
		int x; cin >> x;
		if (x == 1) {
			int i; char c; cin >> i >> c; i--;
			alph[ind[i]].erase(i);
			alph[c - 'a'].insert(i);
			ind[i] = c - 'a';
		}
		else {
			int l, r; cin >> l >> r;
			int ans = 0;
			for (int i = 0; i < 26; ++i)
			{
				if (alph[i].lower_bound(l-1) != alph[i].upper_bound(r-1)) ans++;
			}
			cout << ans << endl;
		}
	}

	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;
typedef long long ll;

int main() {
	int n; cin >> n;
	string s; cin >> s;
	vector<set<int> > alph(26);
	vector<int> ind(n);

	for (int i = 0; i < n; ++i)
	{
		alph[s[i] - 'a'].insert(i);
		ind[i] = s[i] - 'a';
	}

	int q; cin >> q;
	for (int i = 0; i < q; ++i)
	{
		int x; cin >> x;
		if (x == 1) {
			int i; char c; cin >> i >> c; i--;
			alph[ind[i]].erase(i);
			alph[c - 'a'].insert(i);
			ind[i] = c - 'a';
		}
		else {
			int l, r; cin >> l >> r;
			int ans = 0;
			for (int i = 0; i < 26; ++i)
			{
				if (alph[i].lower_bound(l-1) != alph[i].upper_bound(r-1)) ans++;
			}
			cout << ans << endl;
		}
	}

	return 0;
}
```