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

int main(){
	ios::sync_with_stdio(0);
	cout << setprecision(10) << fixed;
	int n, q;
	string s;
	cin >> n >> s >> q;
	set<int> cnt[26];
	for (int i = 0; i < 26; ++i){
		cnt[i].insert(1e9 + 1);
	}
	for (int i = 0; i < n; ++i){
		cnt[s[i] - 'a'].insert(i);
	}
	while (q--){
		int t;
		cin >> t;
		if (t == 1){
			int i;
			char c;
			cin >> i >> c;
			i--;
			cnt[s[i] - 'a'].erase(i);
			cnt[c - 'a'].insert(i);
			s[i] = c;
		}
		else{
			int l, r;
			cin >> l >> r;
			l--, r--;
			int ans = 0;
			for (int i = 0; i < 26; ++i){
				if (cnt[i].size() == 0){
					continue;
				}
				int now = *cnt[i].lower_bound(l);

				if (now >= l && now <= r){
					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;

int main(){
	ios::sync_with_stdio(0);
	cout << setprecision(10) << fixed;
	int n, q;
	string s;
	cin >> n >> s >> q;
	set<int> cnt[26];
	for (int i = 0; i < 26; ++i){
		cnt[i].insert(1e9 + 1);
	}
	for (int i = 0; i < n; ++i){
		cnt[s[i] - 'a'].insert(i);
	}
	while (q--){
		int t;
		cin >> t;
		if (t == 1){
			int i;
			char c;
			cin >> i >> c;
			i--;
			cnt[s[i] - 'a'].erase(i);
			cnt[c - 'a'].insert(i);
			s[i] = c;
		}
		else{
			int l, r;
			cin >> l >> r;
			l--, r--;
			int ans = 0;
			for (int i = 0; i < 26; ++i){
				if (cnt[i].size() == 0){
					continue;
				}
				int now = *cnt[i].lower_bound(l);

				if (now >= l && now <= r){
					ans++;
				}
			}
			cout << ans << "\n";
		}
	}
	return 0;
}


```