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 mp make_pair
#define endl '\n'
typedef long long ll;
 
const int INF=1e9+5;
const ll lim=(ll)3e16;
const int N=3e3+5;
const int mod=1e9+7;
const ll oo=1e18+5;
 

int main(){
	ios::sync_with_stdio(false); cin.tie(NULL);
	int n,q;
	string s;
	cin>>n>>s>>q;
	vector<set<int>> is(26);
	for(int i=0;i<n;++i){
		is[s[i]-'a'].insert(i);
	}
	while(q--){
		int opt;
		cin>>opt;
		if(opt==1){
			int i;
			char c;
			cin>>i>>c;
			--i;
			is[s[i]-'a'].erase(i);
			s[i]=c;
			is[s[i]-'a'].insert(i);
		}
		else{
			int l,r;
			cin>>l>>r;
			--l;
			int cnt=0;
			for(int i=0;i<26;++i){
				auto it=is[i].lower_bound(l);
				if(it!=is[i].end()&&*it<r) ++cnt;
			}
			cout<<cnt<<endl;
		}
	}
	return 0;
}

