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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <set>
using namespace std;
int main(){
    int N, Q;
    cin >> N;
    string S;
    cin >> S >> Q;
    vector<set<int>> alp(26);
    for(int i = 0; i < N; i++){
        int ind = S[i] - 'a';
        alp[ind].insert(i);
    }
    for(int i = 0; i < Q; i++){
        int flag;
        cin >> flag;
        if(flag == 1){
            int iq;
            char c;
            cin >> iq >> c;
            iq--;
            if(S[iq] != c){
                int ind1 = S[iq] - 'a';
                int ind2 = c - 'a';
                alp[ind1].erase(iq);
                alp[ind2].insert(iq);
            }
        }
        else{
            int cnt = 0;
            int lq, rq;
            cin >> lq >> rq;
            lq--;
            rq--;
            for(int j = 0; j < 26; j++){
                if(alp[j].empty()) continue;
                auto ite = alp[j].lower_bound(lq);
                if(ite != alp[j].end() && *ite <= rq) {
                    cnt++;
                }
            }
            cout << cnt << endl;
        }
    }
}