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 <bits/stdc++.h>
using namespace std;

using ll = long long;
#define rep(i,n) for(int i = 0; i < (n); i++)
#define rrep(i,n) for(int i = (n); i > 0; i--)
#define all(n) n.begin(), n.end()
#define print(n) cout << n << endl

int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    
    ll n,q,t,l,r,ans;
    string s;
    char c;
    vector<set<ll>> a(26);
    cin >> n >> s >> q;

    rep(i,s.size()) a[s[i]-'a'].insert(i);
    rep(i,q){
        cin >> t;
        if(t==1){
            cin >> l >> c;
            a[s[l-1]-'a'].erase(l-1);
            a[c-'a'].insert(l-1);
            s[l-1] = c;
        }else{
            cin >> l >> r;
            ans = 0;
            rep(j,26){
                auto itr = a[j].lower_bound(l-1);
                if(itr!=a[j].end() && *itr<r) ans++;
            }
            print(ans);
        }
    }
    return 0;
}