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 ;
typedef long long ll ;
const int mod=1000000007 ;

void solve(){
	
	int n ;
	cin>>n ;

	string s ;
	cin>>s ;

	int q ;
	cin>>q ;

	vector<set<int>> is(26) ;

	for(int i=0;i<n;i++){
		is[s[i]-'a'].insert(i) ;
	}

	for(int i=1;i<=q;i++){

		int a ;
		cin>>a ;

		if(a==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 ans=0 ;

			for(int i=0;i<26;i++){

				auto it=is[i].lower_bound(l);

				if(it!=is[i].end() && *it<r){
					ans++ ;
				}

			}

			cout<<ans<<"\n" ;
		}

	}

}

int main(){
	
	int y ;

	y=1 ;
	//cin>>y ;

	while(y--)
	{	
		solve() ;
	}

}