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>

using namespace std;

int main(void)
{
	int n, q;
	string s;
	cin >> n >> s >> q;

	set<int> st[26];
	for (int i = 0; i < n; i++) {
		st[s[i] - 'a'].insert(i + 1);
	}

	for (int i = 0; i < q; i++) {
		int t;
		cin >> t;
		if (t == 1) {
			int p;
			char c;
			cin >> p >> c;

			st[s[p - 1] - 'a'].erase(p);
			s[p - 1] = c;
			st[c - 'a'].insert(p);
		} else {
			int l, r;
			cin >> l >> r;

			int ans = 0;
			for (int i = 0; i < 26; i++) {
				auto it = st[i].lower_bound(l);
				if (it != st[i].end() && *it <= r) {
					ans++;
				}
			}
			printf("%d\n", ans);
		}
	}

	return 0;
}