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 <bits/stdc++.h>

using namespace std;

int main(int, char**) {
	int n;
	string s;
	int q;
	cin >> n >> s >> q;
	vector<set<int>> alp;
	alp.assign(26, set<int>());

	for (int i=0; i<n; i++) {
		char c = s.at(i);
		alp[s.at(i) - 'a'].insert(i + 1);
	}


	for (int i=0; i<q; i++) {
		int qu;
		scanf("%d", &qu);
		if (qu==1) {
			int l;
			char c;
			scanf("%d %c", &l, &c);
			char b = s.at(l - 1);
			alp[b - 'a'].erase(l);
			s.at(l - 1) = c;
			alp[c - 'a'].insert(l);
		} else {
			int l, r;
			scanf("%d %d", &l, &r);
			int cnt = 0;
			for (int i=0; i<26; i++) {
				if (alp[i].empty()) continue;
				auto itr = alp[i].lower_bound(l);
				if (itr != alp[i].end() && *itr <= r) cnt++;
			}
			cout << cnt << endl;
		}
	}
	return 0;
}
