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
#include "iostream"
#include "vector"
#include "set"
#include "string"

using namespace std;

int main(void) {
	int n, q;
	string s;
	cin >> n >> s >> q;

	vector<set<int>> pos(26);
	for (int i = 0; i < n; ++i) {
		pos.at(s.at(i) - 'a').insert(i);
	}

	for (int tmp_i = 0; tmp_i < q; ++tmp_i) {
		int type;
		cin >> type;
		if (type == 1) {
			int ti;
			char c;
			cin >> ti >> c;
			int i = ti - 1;
			pos.at(s.at(i) - 'a').erase(i);
			pos.at(c - 'a').insert(i);
			s.at(i) = c;
		}
		else {
			int tl, tr;
			cin >> tl >> tr;
			int l = tl - 1, r = tr - 1;
			int num = 0;
			for (int j = 0; j < 26; ++j)
				if (pos.at(j).lower_bound(l) != pos.at(j).end() && *pos.at(j).lower_bound(l) <= r)
					num++;
			cout << num << endl;
		}
	}
}