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;
 
signed main () {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int n;
	cin >> n;
	string s;
	cin >> s;
	vector < set < int > > sts(26);
	for (int i = 0; i < n; ++i) 
		sts[s[i] - 'a'].insert(i);
	int q;
	cin >> q;
	for (int i = 0; i < q; ++i) {
		int t, l, r, p;
		char c;
		cin >> t;
		if (t == 1) {
			cin >> p >> c;
			sts[s[p - 1] - 'a'].erase(p - 1);
			sts[c - 'a'].insert(p - 1);
			s[p - 1] = c;
		}
		else if (t == 2) {
			cin >> l >> r;
			--l; --r;
			int ans = 0;
			for (int j = 0; j < 26; ++j) {
				auto it = sts[j].lower_bound(l);
				if (it != sts[j].end() && *it <= r)
					++ans;
			}
			cout << ans << '\n';
		}
	}
}
