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 <cstdio>
#include <algorithm>
#include <set>
#define N 500005
using namespace std;

int n, m, z, num, cnt, l, r;
char b[5], a[N];
set<int> v[30];
int main()
{
	int i,j, t1, t2;
	cin >> n;
	scanf ("%s", &a[1]);
	cin >> m;
	for (i = 1; i <= n; i++) {
		num = a[i] - 'a';
		v[num].insert(i);
	}
	for (i = 0; i < 26; i++) v[i].insert(n + 100);
	while (m--) {
		scanf ("%d %d", &z,&t1);
		if (z == 1) {
			scanf ("%s", b);
			num = a[t1] - 'a';
			v[num].erase(t1);
			num = b[0] - 'a';
			v[num].insert(t1);
			a[t1] = b[0];
		} else {
			scanf ("%d", &t2);
			cnt = 0;
			for(i = 0; i < 26; i++) {
				l = *(v[i].lower_bound(t1));
				r = *(v[i].upper_bound(t2));
				if (l != r) cnt++;
			}
			printf("%d\n", cnt);
		}
	}
    return 0;
}
