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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <string>
#include <iomanip>
using namespace std;
#define faster       ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef long long ll;
typedef long double ld;

ll n,q;
vector<set<ll> > st;
string str;

void build(ll id=1, ll l=0, ll r=n){
  if(r-l==1){
    st[id].insert(str[l]);
    return;
  }
  ll mid = (l+r)/2;
  build(2*id,l,mid);
  build(2*id+1,mid,r);
  st[id].insert(st[2*id].begin(),st[2*id].end());
  st[id].insert(st[2*id+1].begin(),st[2*id+1].end());
}

set<ll> ts;
void query(ll x, ll y, ll id=1, ll l=0, ll r=n){
  if(x>=r||y<=l){
		return;
	}
	if(x<=l&&y>=r){
    ts.insert(st[id].begin(),st[id].end());
    return;
	}
	ll mid=(l+r)/2;
  query(x,y,2*id,l,mid);
  query(x,y,2*id+1,mid,r);
  return;
}

void update(ll x,char y,ll id=1, ll l=0, ll r = n){
	if((r-l)<2){
    st[id].clear();
		st[id].insert(y);
		return;
	}
	ll mid=(l+r)/2;
	if(x<mid){
		update(x,y,2*id,l,mid);
	}
	else{
		update(x,y,2*id+1,mid,r);
	}
  st[id].clear();
	st[id].insert(st[2*id].begin(),st[2*id].end());
  st[id].insert(st[2*id+1].begin(),st[2*id+1].end());
}


int main(){
  cin >> n;
  cin >> str;
  cin >> q;
  st.resize(4*n);
  build();
  while(q--){
    ll a;
    cin >> a;
    if(a==1){
      ll b;
      char ch;
      cin >> b >> ch;
      update(b,ch);
    }
    else{
      ll c,d;
      cin >> c >> d;
      ts.clear();
      query(c-1,d);
      cout << ts.size() << endl;
    } 
  }
  // for(ll i=0;i<4*n;i++){
  //   cout << st[i].size() << " ";
  // }
  

}