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: Accepted)

#include<bits/stdc++.h>
using namespace std;
set<int> s[26];
int n,m;
string str;
int main(){
    scanf("%d",&n);
    cin>>str;
    for(int i=0;i<n;i++)
        s[str[i]-'a'].insert(i);
    scanf("%d",&m);
    int k;
    for(int i=0;i<m;i++){
        scanf("%d",&k);
        if(k==1){
            int p;char q;
            scanf("%d %c",&p,&q);
            s[str[p-1]-'a'].erase(p-1);
            s[q-'a'].insert(p-1);
            str[p-1]=q;
        }
        else{
            int p,q;
            scanf("%d %d",&p,&q);
            int sum=0;
            p--;q--;
            for(int i=0;i<26;i++)
                if(*s[i].lower_bound(p)<=q&&s[i].lower_bound(p)!=s[i].end())
                    sum++;
            printf("%d\n",sum);
        }
        getchar();

    }


}