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 <set>
using namespace std;

set<int> adj[29];

int main() {
	int n;
	cin >> n;
	string s;
	cin >> s;
	for(int i = 0; i < n; ++i)
		adj[s[i] - 'a'].insert(i + 1);
	int q;
	cin >> q;
	while(q--) {
		int t;
		cin >> t;
		if(t == 1) {
			int left; char c;
			cin >> left >> c;
			for(int i = 0; i < 27; ++i) {
				if(adj[i].find(left) != adj[i].end()) {
					adj[i].erase(left);
				}
			}
			adj[c - 'a'].insert(left);
		}
		else {
			int left, right, answer = 0;
			scanf("%d%d", &left, &right);
			set<int> :: iterator it;
			for(int i = 0; i < 27; ++i) {
				it = adj[i].lower_bound(left);
				answer += (it != adj[i].end() && *it <= right && *it >= left);
			}
			cout << answer << '\n';
		}
	}
}
