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 <cstdio>
#include <cstring>
#include <cmath>
#include <utility>
#include <iostream>
#include <algorithm>
#include <vector>
#include <forward_list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#define rep(i, s, g) for ((i) = (s); (i) < (g); ++(i))
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
const ll MOD = 1e9 + 7;
const ll INF = (1ll << 62) - 1;

int main(void)
{
    int n, q;
    string s;
    int i, j;
    set<int> alphabet[30];
    int x, y, z;
    char c;

    cin >> n >> s >> q;

    rep(i, 0, n)
    {
        alphabet[s[i] - 'a'].insert(i);
    }

    rep(i, 0, q)
    {
        cin >> x >> y;

        y--;

        if (x == 1)
        {
            cin >> c;
        }
        else
        {
            cin >> z;
            z--;
        }

        if (x == 1)
        {
            alphabet[s[y] - 'a'].erase(y);
            s[y] = c;
            alphabet[c - 'a'].insert(y);
        }
        else
        {
            int ans = 0;

            rep(j, 0, 26)
            {
                auto it = alphabet[j].lower_bound(y);

                if (z >= *it && it != alphabet[j].end())
                {
                    ans++;
                }
            }

            printf("%d\n", ans);
        }
    }
}