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 N,Q;
	string S;
	cin >>N >> S >> Q;
	vector<set<int>>Set(26);
	for (int n = 0;n<S.size();++n) {
		Set[S[n] - 'a'].insert(n);
	}
	vector<int>Ans;
	for (int n = 0;n<Q; ++n) {
		int q1;
		cin >> q1;
		if (1 == q1) {
			int q2;
			char q3;
			cin >> q2 >> q3; 
			Set[S[q2 - 1]-'a'].erase(q2 - 1); 
			S[q2 - 1] = q3;
			Set[S[q2 - 1]-'a'].insert(q2 - 1);
		}
		else
		{
			int q2, q3;
			cin >> q2 >> q3;
			int ans = 0;
			for (int m = 0;m < 26;++m) {
				set<int>::iterator it = Set[m].lower_bound(q2-1);
				if (it != Set[m].end() && *it <q3) {
					++ans;
				}
			}
			Ans.push_back(ans);
		}
	}
	for (int n = 0;n<Ans.size();++n) {
		cout << Ans[n] << endl;
	}
	return 0;
}