#include <bits/stdc++.h>

using namespace std;


int main() {
	int n;
	cin >> n;
	string s;
	cin >> s;

	vector<set<int>> freq(26);

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

	int q;
	cin >> q;

	for (int i = 0; i < q; i++) {
		int type;
		cin >> type;
		if (type == 1) {
			int index;
			char ch;
			cin >> index >> ch;
			index--;
			char oldCharacter = s[index];
			s[index] = ch;
			freq[oldCharacter - 'a'].erase(index);
			freq[ch - 'a'].insert(index);
		}
		else {
			int left, right;
			cin >> left >> right;
			left--, right--;
			int res = 0;
			for (int j = 0; j < 26; j++) {
				auto it = freq[j].lower_bound(left);
				if (it == freq[j].end()) continue;
				int val = *it;
				if (val >= left && val <= right) res++;
			}

			cout << res << endl;
		}
	}
}
### Prompt

Create a solution in CPP for 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;

	vector<set<int>> freq(26);

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

	int q;
	cin >> q;

	for (int i = 0; i < q; i++) {
		int type;
		cin >> type;
		if (type == 1) {
			int index;
			char ch;
			cin >> index >> ch;
			index--;
			char oldCharacter = s[index];
			s[index] = ch;
			freq[oldCharacter - 'a'].erase(index);
			freq[ch - 'a'].insert(index);
		}
		else {
			int left, right;
			cin >> left >> right;
			left--, right--;
			int res = 0;
			for (int j = 0; j < 26; j++) {
				auto it = freq[j].lower_bound(left);
				if (it == freq[j].end()) continue;
				int val = *it;
				if (val >= left && val <= right) res++;
			}

			cout << res << endl;
		}
	}
}
```