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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <queue>
#include <tuple>
#include <set>
#include <map>

using namespace std;

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

    vector< set<int> > vst(26);
    for(int i=0; i<N; i++) {
        vst[S[i] - 'a'].insert(i);
    }

    while(Q--) {
        int q;
        cin >> q;
        if(q == 1) {
            int i;
            char c;
            cin >> i >> c;
            --i;
            vst[S[i] - 'a'].erase(i);
            S[i] = c;
            vst[S[i] - 'a'].insert(i);
        } else if(q == 2) {
            int l, r;
            cin >> l >> r;
            --l; --r;
            int ans = 0;
            for(int i=0; i<26; i++) {
                if(vst[i].upper_bound(r) != vst[i].lower_bound(l)){
                    ++ans;
                }
            }
            cout << ans << endl;
        }
    }
    return 0;
}