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 modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000

int main(){
	
	int N;
	cin>>N;
	string S;
	cin>>S;
	
	int Q;
	cin>>Q;
	
	vector<set<int>> s(26);
	for(int i=0;i<S.size();i++){
		s[S[i]-'a'].insert(i);
	}
	
	for(int i=0;i<Q;i++){
		int t;
		cin>>t;
		if(t==1){
			int ind;
			char c;
			cin>>ind>>c;
			ind--;
			s[S[ind]-'a'].erase(ind);
			S[ind] = c;
			s[S[ind]-'a'].insert(ind);
		}
		else{
			int l,r;
			cin>>l>>r;
			l--;r--;
			int ans = 0;
			for(int j=0;j<26;j++){
				auto it= s[j].lower_bound(l);
				if(it==s[j].end())continue;
				if((*it)<=r)ans++;
			}
			cout<<ans<<endl;
		}
	}
	
	return 0;
}
