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: Time Limit Exceeded)

#include <algorithm>
#include <cmath>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <set>
#include <stack>
#include <string>
#include <unordered_set>
#include <vector>

using namespace std;
using LLONG = long long;
const LLONG MOD = 1000000007;

string S;
vector<set<int>> posLists(26);

void Type1(const int i, const char ch)
{
    int prevCh = S[i - 1] - 'a';
    int newCh = ch - 'a';
    posLists[prevCh].erase(i);
    posLists[newCh].insert(i);
    S[i - 1] = ch;
}

const int Type2(const int L, const int R)
{
    int ret = 0;
    // 各文字について[L, R]の範囲に存在するかどうか調べる:O(logN)
    for (const auto &posList : posLists)
    {
        // L以上の要素を持つかどうか探索する
        const auto lb = lower_bound(posList.begin(), posList.end(), L);
        if (lb != posList.end())
        {
            // L以上の要素のうち，最小の要素がR以下であればOK
            const int lbNum = *lb;
            if (lbNum <= R) ++ret;
        }
    }
    return ret;
}

int main()
{
    int N; cin >> N >> S;
    for (int i = 0; i < N; ++i)
    {
        int ch = static_cast<int>(S[i] - 'a');
        posLists[ch].insert(i + 1);
    }
    int Q; cin >> Q;
    // Type1はO(1)，Type2はO(log(N))より，総計算量はO(Qlog(N))
    for (int q = 0; q < Q; ++q)
    {
        int type; cin >> type;
        if (type == 1)
        {
            int iq; char cq; cin >> iq >> cq;
            Type1(iq, cq);
        }
        else
        {
            int lq, rq; cin >> lq >> rq;
            cout << Type2(lq, rq) << endl;
        }
    }
}