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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define N 2000000
bool tree[N][26];
string s;
bool ans[26];
void build(int st,int en,int node)
{
	if(st==en)
	{
		for(int i=0;i<26;++i)tree[node][i]=false;
		int x=s[st]-'a';
		tree[node][x]=true;
	    return;
	}
	int mid=(st+en)/2;
	build(st,mid,2*node);
	build(mid+1,en,2*node+1);
	for(int i=0;i<26;++i)tree[node][i]=tree[2*node][i]||tree[2*node+1][i];
}
void update(int st,int en,int x,int node)
{
	if(x<st||x>en||st>en)return;
	if(st==en&&en==x)
	{
		for(int i=0;i<26;++i)tree[node][i]=false;
		int x=s[st]-'a';
	    tree[node][x]=true;
	    return;
	}
	int mid=(st+en)/2;
	if(mid>=x)build(st,mid,2*node);
	else build(mid+1,en,2*node+1);
	for(int i=0;i<26;++i)tree[node][i]=tree[2*node][i]||tree[2*node+1][i];
}
void query(int st,int en,int l,int r,int node)
{
	if(st>en||st>r||en<l)return;
	if(st>=l&&en<=r)
	{
		for(int i=0;i<26;++i)ans[i]=ans[i]||tree[node][i];
		return;
	}
	int mid=(st+en)/2;
	query(st,mid,l,r,2*node);
	query(mid+1,r,l,r,2*node+1);
}
int main() {
	// your code goes here
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n; cin>>n; cin>>s;
	int q; cin>>q;
	memset(tree,0,sizeof(tree));
	build(0,n-1,1);
	while(q--)
	{
		int x,y,z; char ch;
		cin>>x;
		if(x==1)
		{
			cin>>y>>ch;
			y--;
			if(s[y]!=ch){ s[y]=ch;
			update(0,n-1,y,1);
		  }
		}
		else
		{
			cin>>y>>z;
			y--,z--;
			for(int i=0;i<26;++i)ans[i]=false;
			query(0,n-1,y,z,1);
			int cnt=0;
			for(int i=0;i<26;++i)if(ans[i])cnt++;
			cout<<cnt<<endl;
		}
	}

}