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;
#define rep(i,n) for (int i = 0; i < (n); ++i)

int main() {
	int N,Q;
	string S;
	vector<set<int>> m(26);

	cin >> N >> S >> Q;
	rep(i,N)
		m.at(S.at(i)-'a').insert(i+1);

	rep(q,Q) {
		int t,i;
		cin >> t >> i;

		if(t == 1) {
			char c;
			cin >> c;
			
			m[S.at(i-1)-'a'].erase(i);
			m[c-'a'].insert(i);
			S.at(i-1) = c;
		} else {
			int r,c=0;
			cin >> r;

			rep(j,26) {
				auto itr_l = m.at(j).lower_bound(i);
				if(itr_l != m.at(j).end()) {
					auto itr_r = m.at(j).upper_bound(r);
					c += (itr_l != itr_r);
				}
			}
			cout << c << endl;
		}
	}
}