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;
string s;
int n;
int c[500001][26];

inline int lowbit(int x){
	return x&(-x);
}

inline void add(int p,int k){
	while(p<=n){
		++c[p][k];
		p+=lowbit(p);
	}
}
inline void amend(int p,int k,int f){
	while(p<=n){
		++c[p][k];
		--c[p][f];
		p+=lowbit(p);
	}
}

inline int sum(int p,int k){
	int ans=0;
	while(p){
		ans+=c[p][k];
		p-=lowbit(p);
	}
	return ans;
} 
inline int ask(int l,int r){
	int ss=0;
	for(int i=0;i<26;++i)if(sum(r,i)-sum(l-1,i))++ss;
	return ss;
}

int main(){
	memset(c,0,sizeof(c));
	register int m,f,a,b;
	register char c;
	cin>>n>>s>>m;
	s=' '+s;
	int i;
	for(i=1;i<=n;++i)add(i,s[i]-'a');
	while(m--){
		cin>>f>>a;
		if(f==1){
			cin>>c;
			amend(a,c-'a',s[a]-'a');
			s[a]=c;
		}
		else{
			cin>>b;
			cout<<ask(a,b)<<endl;
		}
	}
}