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 <string>
#include <set>
#include <vector>

using namespace std;

int main() {
	ios_base::sync_with_stdio(false);
	int n;
	cin>>n;
	string s;
	cin>>s;
	vector<set<int>> v(26);
	for (int i = 0; i < s.size(); i++) {
		v[s[i]-'a'].insert(i+1);
	}
	int q;
	cin>>q;
	for (int i = 0; i < q; i++) {
		int f;
		cin>>f;
		if (f == 1) {
			int j;
			char c;
			cin>>j>>c;
			if (s[j-1] != c) {
				v[s[j-1]-'a'].erase(j);
				v[c-'a'].insert(j);
				s[j-1] = c;
			}
		} else {
			int a, b;
			cin>>a>>b;
			int res = 0;
			for (int k = 0; k < 26; k++) {
				auto ia = v[k].lower_bound(a);
				auto ib = v[k].upper_bound(b);
				if (ia != ib) res++;
			}
			cout<<res<<endl;
		}
	}
	return 0;
}