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;

typedef long long ll;

const int N = 5e5 + 5;
const int ALPHA = 26;

int n;
string s;

set<int> pos[ALPHA];

int main() {
	cin >> n >> s;
	for (int i = 0; i < n; i++) {
		pos[s[i] - 'a'].insert(i);
	}
	int q;
	cin >> q;
	while (q--) {
		int typ;
		cin >> typ;
		if (typ == 1) {
			int idx;
			char c;
			cin >> idx >> c;
			idx--;
			pos[s[idx] - 'a'].erase(idx);
			pos[c - 'a'].insert(idx);
			s[idx] = c;
		} else {
			int L, R, res = 0;
			cin >> L >> R;
			L--, R--;
			for (int i = 0; i < ALPHA; i++) {
				auto it = pos[i].lower_bound(L);
				if (it != pos[i].end() && *it <= R)
					res++;
			}
			cout << res << "\n";
		}
	}
	return 0;
}
