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;
const int MX=5e5+4;
int bit[MX][27];
void upd(int x,int id,int vl)
{
	while(x<MX)
	{
		bit[x][id]+=vl;
		x+=x&-x;
	}
}
int get(int x,int id)
{
	int ans=0;
	while(x>0)
	{
		ans+=bit[x][id];
		x-=x&-x;
	}
	return ans;
}
int main()
{
	string s;
	int n,q;
	cin>>n>>s;
	s="#"+s;
	for(int i=1;i<=n;i++){
		upd(i,s[i]-'a',1);
	}
	cin>>q;
	while(q--)
	{
		char chr;
		int ch,l,r;
		cin>>ch;
		if(ch==1)
		{
			cin>>l>>chr;
			upd(l,s[l]-'a',-1);
			upd(l,chr-'a',1);
			s[l]=chr;
		}
		else
		{
			int res=0;
			cin>>l>>r;
			for(int j=0;j<26;j++)
			{
				if(get(r,j)-get(l-1,j)>0) res++;
			}
			cout<<res<<endl;
		}
	}
	return 0;
}