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>
#define int long long
using namespace std;

signed main(){
	int N; string S; cin>>N>>S;
	map<char,set<int>> dic;
	for(char i='a';i<='z';i++){
		set<int> tmp;
		dic[i]=tmp;
	}
	for(int i=0;i<N;i++){
		dic[S[i]].insert(i);
	}
	int Q; cin>>Q;
	for(int i=0;i<Q;i++){
		int a; cin>>a;
		if(a==1){
			int i; char c; cin>>i>>c; i--;
			if(S[i]!=c){
				dic[S[i]].erase(i);
				dic[c].insert(i);
				S[i]=c;
			}
		}
		else if(a==2){
			int l,r; cin>>l>>r; l--; r--;
			int ans=0;
			for(char p='a';p<='z';p++){
				if(dic[p].size()==0)continue;
				int tmp=*dic[p].lower_bound(l);
				if(l<=tmp && tmp<=r && dic[p].lower_bound(l)!=dic[p].end())ans++;
			}
			cout<<ans<<endl;
		}
	}
	return 0;
}