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;

const int MAX = 5e5 + 5;
int n, m;
char s[MAX];
set <int> pos[26];

int main() {
	scanf("%d%s%d", &n, s + 1, &m);

	for(int i = 1; i <= n; i++)
		pos[s[i] - 'a'].insert(i);

	while(m--) {
		int op; scanf("%d", &op);

		if(op == 1) {
			int p; char w[5];
			scanf("%d%s", &p, w);

			if(s[p] == w[0])
				continue;

			auto &st = pos[s[p] - 'a'];

			assert(st.find(p) != st.end());
			st.erase(st.find(p));

			auto &nst = pos[w[0] - 'a'];
			nst.insert(p);

			s[p] = w[0];
		}

		else {
			int l, r; scanf("%d%d", &l, &r);

			int ans = 0;
			for(int i = 0; i < 26; i++) {
				auto it = pos[i].lower_bound(l);

				if(it != pos[i].end() && *it <= r)
					ans++;
			}

			printf("%d\n", ans);
		}
	}

	return 0;
}