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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define zero_pad(num) setfill('0') << std::right << setw(num)
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<int, int>;

int main() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    vector<set<int>> v(26);
    rep(i, n){
        v[s[i] - 'a'].insert(i);
    }

    int q;
    cin >> q;
    rep(i, q){
        int t;
        cin >> t;
        if(t == 1){
            int k;
            char c;
            cin >> k >> c;
            k--;
            v[s[i] - 'a'].erase(k);
            v[c - 'a'].insert(k);
            s[i] = c;
        }else{
            int l, r;
            cin >> l >> r;
            l--;
            r--;
            int cnt = 0;
            rep(j, 26){
                auto it = v[j].lower_bound(l);
                if(it != v[j].end()){
                    if(*it <= r)cnt++;
                }
            }
            cout << cnt << endl;
        }
    }
    
}