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;
//BEGIN CUT HERE
template <typename T,typename E>
struct SegmentTree{
  typedef function<T(T,T)> F;
  typedef function<T(T,E)> G;
  int n;
  F f;
  G g;
  T d1;
  E d0;
  vector<T> dat;
  SegmentTree(){};
  SegmentTree(int n_,F f,G g,T d1,
	      vector<T> v=vector<T>()):
    f(f),g(g),d1(d1){
    init(n_);
    if(n_==(int)v.size()) build(n_,v);
  }
  void init(int n_){
    n=1;
    while(n<n_) n*=2;
    dat.clear();
    dat.resize(2*n-1,d1);
  }
  void build(int n_, vector<T> v){
    for(int i=0;i<n_;i++) dat[i+n-1]=v[i];
    for(int i=n-2;i>=0;i--)
      dat[i]=f(dat[i*2+1],dat[i*2+2]);
  }
  void update(int k,E a){
    k+=n-1;
    dat[k]=g(dat[k],a);
    while(k>0){
      k=(k-1)/2;
      dat[k]=f(dat[k*2+1],dat[k*2+2]);
    }
  }
  inline T query(int a,int b){
    T vl=d1,vr=d1;
    for(int l=a+n,r=b+n;l<r;l>>=1,r>>=1) {
      if(l&1) vl=f(vl,dat[(l++)-1]);
      if(r&1) vr=f(dat[(--r)-1],vr);
    }
    return f(vl,vr);
  }
  
};
//END CUT HERE
signed main(){
    int n;
    cin >> n;
    SegmentTree<int,int> seg(n,
                [](int a,int b){return a | b;},
                [](int a,int b){return b;},
                0);
    string S;
    cin >> S;
    for (int i = 0; i < n; i++){
        seg.update(i, 1 << (S[i] - 'a'));
    }
    int q;
    cin >> q;
    for(int i=0;i<q;i++){
        int a;
        cin >> a;
        if (a == 1){
            int j;
            char c;
            cin >> j >> c;
            j--;
            seg.update(j, 1 << (c - 'a'));
        } else {
            int l, r;
            cin >> l >> r;
            l--;
            cout << __builtin_popcount(seg.query(l, r)) << endl;
        }
    }
    return 0;
}