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 N = 5e5 + 5;
char s[N];
set<int> st[30];

int main()
{
	int n, q;
	scanf("%d%s%d", &n, s, &q);	
	for (int i = 0; i < n; i++) {
		st[s[i] - 'a'].insert(i + 1);
	}
	while (q--) {
		int op;
		scanf("%d", &op);
		if (op == 1) {
			int pos; char ch;
			scanf("%d %c", &pos, &ch);
			for (int i = 0; i < 26; i++) {
				if (st[i].size() == 0) continue;
				if (st[i].find(pos) != st[i].end()) {
					st[i].erase(pos);
					break;
				}
			}
			st[ch - 'a'].insert(pos);
		} else {
			int l, r, ans = 0;
			scanf("%d%d", &l, &r);
			for (int i = 0; i < 26; i++) {
				if (st[i].size() == 0) continue;
				auto it = st[i].lower_bound(l);
				if (it != st[i].end() && *it <= r) ans++;
			}
			printf("%d\n", ans);
		}
	}
	return 0;
}