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)

//#define _GLIBCXX_DEBUG
#include<bits/stdc++.h>

using namespace std;


void solve() {
    int N, Q; string S;
    cin >> N >> S >> Q;

    vector<set<int>> s(26);

    for (int i = 0; i < (int)S.size(); i++) {
        s[S[i] - 'a'].insert(i+1);
    }

    for (int q = 0; q < Q; q++) {
        int query; cin >> query;

        if (query == 1) {
            int i; char c; 
            cin >> i >> c;
            s[S[i] - 'a'].erase(i);
            s[c - 'a'].insert(i);
            S[i] = c;
        }

        else if (query == 2) {
            int l, r; cin >> l >> r;

            int cnt = 0;

            for (int i = 0; i < 26; i++) {
                if (s[i].size() != 0) {
                    auto lit = s[i].lower_bound(l);
                    auto rit = s[i].upper_bound(r);
                    if (lit != rit) cnt++;
                }
            }

            cout << cnt << endl;
        }
    }
}


int main() {
    solve();
    return 0;
}