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++ (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<stack>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
#define me(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
const int N=500005;
int x,n,i,m,q,l,r,j,sum;
char s[N],ch;
int main()
{
	set<int> a[27];
	set<int>::iterator it;
	cin>>n;
	getchar();
	gets(s+1);
	for(i=1;s[i]!='\0';i++)
	{
		a[s[i]-'a'].insert(i);
	}
	cin>>m;
	while(m--)
	{
		cin>>q;
		if(q==1)
		{
			cin>>x>>ch;
			a[s[x]-'a'].erase(x);
			a[ch-'a'].insert(x);
			s[x]=ch;
		}
		else 
		{
			sum=0;
			cin>>l>>r;
			for(i=l;i<=r;i++)
			{
				for(j=0;j<26;j++)
				{
					auto p=a[j].lower_bound(l);
					if(p!=a[j].end()&&*p<=r)sum++;
				}
			}
			cout<<sum<<endl;
		}
	}
	return 0;
}