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;

int main(){
	int n,q;
	string s;
	cin>>n>>s>>q;
	s=" "+s;
	
	vector<set<int>>Si(26);
	// 各アルファベットごとに、出現位置をsetで記録していく。
	for(int pos=1; pos<=n; pos++){
		int alp=s[pos]-'a';
		Si[alp].insert(pos);
	}

	for(int qi=0;qi<q;qi++){
		int type;	cin>>type;

		if(type==1){
			int pos;	char c;
			cin>>pos>>c;

			int alp=s[pos]-'a';
			Si[alp].erase(pos);
			s[pos]=c;		alp=s[pos]-'a';
			Si[alp].insert(pos);
		}else{
			int L,R;	cin>>L>>R;

			int ans=0;	// L-R 範囲内の、アルファベットの種類数
			for(int alp=0; alp<26; alp++){
				auto it=Si[alp].lower_bound(L);
				// it: L 以降で、アルファベット（alp）が出る最小のポジション

				if(it != Si[alp].end() && *it<=R ){
					// it が Rより左にあるなら、範囲内にアルファベット（alp）が存在するよね！
					ans++;
				}
			}
			cout<<ans<<endl;
		}
	}


	return 0;
}
