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 forr(i,a,n)	for(int i=a; i<n; i++)
#define loop(i,a,n)	for(int i=a; i>=n; i--)

int main() 
{	
	ios_base :: sync_with_stdio(false),cin.tie(NULL),cout.tie(0);
	
	int a;
	cin>>a;
	string s;
	cin>>s;
	int b;
	cin>>b;
	vector<set<int>> v(26);
	map<int,char> m;
	forr(i,0,s.size())
	{
		v[int(s[i]-'a')].insert(i);
		m[i]=s[i];
	}
	while(b--)
	{
		int x,y;
		cin>>x>>y;
		if(x==1)
		{
			char z;
			cin>>z;
			char t=m[y-1];
			v[int(t-'a')].erase(y-1);
			v[int(z-'a')].insert(y-1);
			s[y-1]=z;
			m[y-1]=z;
		}	
		else
		{
			int z,ct=0;
			cin>>z;
			forr(i,0,26)
			{
				auto it=v[i].lower_bound(y-1);
				if(it!=v[i].end() and *it<z)	ct++;
			}
			cout<<ct<<'\n';
		}
	}
	return 0;	
}