Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7

Sample Output 1

3
1
5

In the first query, cdbb contains three kinds of letters: b , c , and d, so we print 3.

In the second query, S is modified to abcdzbd.

In the third query, a contains one kind of letter: a, so we print 1.

In the fourth query, S is modified to abcazbd.

In the fifth query, S does not change and is still abcazbd.

In the sixth query, abcazbd contains five kinds of letters: a, b, c, d, and z, so we print 5.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/*
	Name:
	Author: xiaruize
	Date:
*/

#include<bits/stdc++.h>
using namespace std;
#define ull unsigned long long
#define MOD 1000000007
#define ALL(a) (a).begin(), (a).end()
#define forn(i, n) for (int i = 0; i < int(n); i++)
#define ll long long
#define form(i,j,n) for(int i=int(j);i<=int(n);i++)
#define pb push_back
#define mk make_pair
#define pii pair<int,int>
#define pis pair<int,string>
#define sec second

inline int read()
{
    char ch=getchar();
    int x=0,f=1;
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+(ch^48),ch=getchar();
    return x*f;
}

int n,tree[26][500005];

int lowbit(int x)
{
	return x&(-x);
 } 
 
void add(int x,int k,int tr)
{
	while(x<=n)
	{
		tree[tr][x]+=k;
		x+=lowbit(x);
	}
}

int sum(int x,int tr)
{
	int ans=0;
	while(x>0)
	{
		if(tree[tr][x])
		{
			ans++;
		}
		x-=lowbit(x);
	}
	return ans;
}

signed main()
{
 	ios::sync_with_stdio(false);
	cin.tie(0);
 	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	cin>>n;
	string s;
	cin>>s;
	for(int i=0;i<n;i++)
	{
		add(i+1,1,s[i]-'a');
	}
	int t;
	cin>>t;
	int x,y,z;
	char c;
	while(t--)
	{
		cin>>x;
		if(x==1)
		{
			cin>>y>>c;
			add(y,-1,s[y-1]-'a');
			add(y,1,c-'a');
			s[y-1]=c;
		}
		else if(x==2)
		{
			cin>>y>>z;
			int ans=0;
			for(int i=0;i<26;i++)
			{
				if(sum(z,i)>sum(y-1,i))
				{
					ans++;
				}
			}
			cout<<ans<<endl;
		}
	}
	return 0;
}