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: Time Limit Exceeded)

#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;
#define x first
#define y second
#define pb push_back
#define ppb pop_back
#define pf push_front
#define ppf pop_front
#define ll long long
#define ld long double
#define ii pair<int,int>
#define ull unsigned long long
#define all(k) k.begin(),k.end()
#define rall(k) k.rbegin(),k.rend()
#define db(x) cerr<<#x<<" = "<<x<<endl
#define len(v) ((int)v.size())
#define endl '\n'




void solve(){
  int n,q;
  string s;
  cin>>n>>s>>q;
  while(q--){
    int k;
    cin>>k;
    if(k==1){
      int i;
      char c;
      cin>>i>>c;
      s[i-1]=c;
    }else{
      int l, r;
      cin>>l>>r;
      set<char> ss;
      for(int i=l;i<=r;i++) ss.insert(s[i-1]);
      cout<<ss.size()<<endl;
    }
  }


}
int32_t main(){
  ios::sync_with_stdio(0); cin.tie(0);
  int t; 
#ifdef TESTCASE
  cin>>t;
#else 
  t = 1;
#endif
  while(t--) solve();
}