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;

class Solution {
public:
	void solve() {
		int n, q;
		string s;
		cin >> n >> s >> q;
		vector<set<int>> idxs(26);
		for(int i=0; i < n; i++)idxs[s[i]-'a'].insert(i);
		while(q--){
			int T;
			cin >> T;
			if(T == 1){
				int i;
				char c;
				cin >> i >> c;
				i--;
				if(s[i] == c)continue;
				idxs[s[i]-'a'].erase(i);
				s[i] = c;
				idxs[s[i]-'a'].insert(i);
			}else{
				int l, r;
				cin >> l >> r;
				l--, r--;
				int ans = 0;
				for(int i=0; i < 26; i++){
					auto it = idxs[i].lower_bound(l);
					if(it != idxs[i].end() && *it <= r)ans++;
				}
				cout << ans << "\n";
			}
		}
		return;
	};
};

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	Solution solution;
	solution.solve();
	return 0;
}