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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int i=0;i<n;++i)
#define rep2(i,i0,n) for(int i=i0;i<n;++i)

int main()
{
    int n,q;
    string s;
    cin >> n >> s >> q;

    vector<set<int>> p(26);
    rep(i,n) {
        int j = s[i]-'a';
        p[j].insert(i);
    }

    vector<int> ans;
    rep(i,q) {
        int t;
        cin >> t;
        if(t == 1) {
            int iq;
            string cq;
            cin >> iq >> cq;
            if(s[iq-1] != cq[0]) {
                p[int(s[iq-1]-'a')].erase(iq-1);
                p[int(cq[0]-'a')].insert(iq-1);
            }           
        }
        else {
            int l,r;
            cin >> l >> r;
            int cnt = 0;
            rep(j,26) {
                auto itr_l = p[j].lower_bound(l-1);
                if(itr_l != p[j].end() && *itr_l < r) ++cnt;
            }
            ans.push_back(cnt);
        }
    }

    for(auto cnt:ans) cout << cnt << endl;
    return 0;
}