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 <bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define mod 1000000007
#define all(v)  v.begin(),v.end()
#define pb push_back
#define size(v)  (int) v.size()
#define fast ios_base:: sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL)
ll power_mod(ll a,ll x){if(x==0) return 1;ll y=power_mod(a,x/2);ll ans=(y*y)%mod; if(x%2) ans=(ans*a)%mod;return ans;}
ll inv(ll a){return power_mod(a,mod-2);}    
ll power(ll a, ll x){ if(x==0) return 1; ll y= power(a,x/2); ll ans=(y*y); if(x%2) ans*=a; return ans;}


int main(){
   fast;

   int n;
   cin>>n;
   string s;
   cin>>s;
   int q;
   cin>>q;
   set<int> c[26];
   for(int i=0;i<n;i++){
        c[s[i]-'a'].insert(i);
   }
   while(q--){
      int t;
      cin>>t;
      if(t==1){
         int i;char x;
         cin>>i>>x;
         i--;
         c[s[i]-'a'].erase(i);
         c[x-'a'].insert(i);
         s[i]=x;
      }
      else{
         int l,r;
         cin>>l>>r;
         l--;r--;
         int count = 0;
         for(int i=0;i<26;i++){
             auto it_lo = lower_bound(all(c[i]),l);
             if(it_lo==c[i].end()) continue;
             auto it_hi = lower_bound(all(c[i]),r);
           //  cout<<*(it_lo)<<" "<<*(it_hi)<<endl;
             if(*(it_hi)==*(it_lo)){
                 if(*(it_hi)<=r and *(it_hi)>=l){
               //    cout<<i<<endl;
                   count++;
                 }
             }
             else{
            //  cout<<i<<endl;;
               count++;
             }
         }
           
             cout<<count<<endl;
      }
   }

   return 0;
}