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;
typedef long long ll;
typedef pair<int,int> P;
#define rep(i,m,n) for(int i=m;i<n;i++)
ll mod=1e9+7;

int main(){
  int n,q;
  string s;
  cin>>n>>s>>q;
  set<int> a[26];
  rep(i,0,n) a[(int)s[i]-97].insert(i);
  
  rep(i,0,q){
    int b;
    cin>>b;
    if(b==1){
      int c;
      char d;
      cin>>c>>d;
      c--;
      rep(j,0,26){
        if(a[j].count(c)) a[j].erase(c);
      }
      a[(int)d-97].insert(c);
    }
    else{
      int l,r;
      cin>>l>>r;
      l--;
      r--;
      int ans=0;
      rep(j,0,26){
        if(a[j].size()){
          int e=*rbegin(a[j]);
          if(e>=l){
            int f=*a[j].lower_bound(l);
            if(f<=r) ans++;
          }
        }
      }
      cout<<ans<<endl;
    }
  }
}