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;
	cin >> n;
	string s;
	cin >> s;
	vector<set<int>> poss(26);
	for (int i = 0; i < n; ++i) 
	{
		poss[s[i] - 'a'].insert(i);
	}
	
	int q;
	cin >> q;
	while(q--) 
	{
		int type;
		cin >> type;
		if (type == 1) 
		{
			int pos;
			char c;
			cin >> pos >> c;
			--pos;
			poss[s[pos] - 'a'].erase(pos);
			s[pos] = c;
			poss[s[pos] - 'a'].insert(pos);
		} 
		else 
		{
			int l, r;
			cin >> l >> r;
			--l, --r;
			int cnt = 0;
			for (int c = 0; c < 26; ++c) 
			{
				auto it = poss[c].lower_bound(l);
				if (it != poss[c].end() && *it <= r) ++cnt;
			}
			cout << cnt << "\n";
		}
	}

	
	return 0;
}
