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;

#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define rep1(i, n) for (ll i = 1; i <= (ll)(n); i++)
#define INF 10000000000
#define MOD 1000000007
using ll = long long;
using Graph = vector<vector<int>>;

int main(){
    int N,Q; cin>>N;
    string S; cin>>S;
    cin>>Q;
    map<char,set<int>> M;
    rep(i,N){
        M[S.at(i)].insert(i);
    }
    rep(i,Q){
        int q; cin>>q;
        if(q==1){
            int i;
            char c;
            cin>>i>>c;
            i--;
            M[S.at(i)].erase(i);
            M[c].insert(i);
        }else{
            int res = 0;
            int l,r; cin>>l>>r;
            l--;
            r--;
            for(auto p:M){
                auto iter = lower_bound(p.second.begin(),p.second.end(),l-1);
                if(iter!=p.second.end()&&(*iter<=r))res++;
            }
            cout<<res<<endl;
        }
    }
}