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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
using namespace std;
template <typename F,typename T>
struct SegTree{
  // 二項演算merge,単位元identify
  T identity;
  F merge; 
  int size;
  vector<T> dat;
  
  // 二項演算fと単位元idを渡して宣言する
  SegTree(F f,T id):merge(f),identity(id){}

  // データの要素の数nを渡して初期化、sizeはnより大きい2の冪乗
  void init(int n){
    size = 1;
    while(size<=n) size *= 2;
    dat.resize(size*2-1,identity);
  }

  // 配列を渡して0(n)で初期化
  void build(vector<T> vec){
    rep(i,vec.size()) dat[size-1+i] = vec[i];
    dfs(0);
  }
  
  T dfs(int k){
    if(k>=size-1) return dat[k];
    else return dat[k] = merge(dfs(2*k+1),dfs(2*k+2));
  }

  // index kの要素をaに変更
  void update(int k,T a){
    k += size - 1;
    dat[k] = a;
    while(k > 0){
      k = (k-1)/2;
      dat[k] = merge(dat[2*k+1],dat[2*k+2]);
    }
  }

  // 区間[a,b)に対するクエリに答える。(k,l,r)=(0,0,size)
  T query(int a,int b,int k,int l,int r){
    if(r<=a||b<=l) return identity;
    
    if(a<=l&&r<=b) return dat[k]; 
    else return merge(query(a,b,2*k+1,l,(l+r)/2),query(a,b,2*k+2,(l+r)/2,r));
  }
  
  void show(){
    int index = 0;
    int num = 1;
    while(index<size){
      rep(i,num){
	if(dat[i+index]==identity) cout << "e ";
	else cout << dat[i+index] << " ";
      }
      cout << "\n";
      num *= 2;
      index = index*2+1;
    }
  }
  
};

int main()
{
  int n,q;
  string s;
  cin >> n >> s >> q;
  vector<int> c(q),x(q),y(q);
  rep(i,q){
    cin >> c[i];
    if(c[i]==1){
      char cc;
      cin >> x[i] >> cc;
      x[i]--;
      int yy = cc - 'a';
      y[i] = yy;
    }
    else{
      cin >> x[i] >> y[i];
      x[i]--;
    }
  }

  auto f = [&](int a,int b){return a|b;};
  int id=0;
  vector<int> vec(n);
  rep(i,n){
    int kk=(int)s[i]-'a';
    vec[i] = 1<<kk;
  }
  SegTree<decltype(f),int> seg(f,id);
  seg.init(n);
  seg.build(vec);
  rep(i,q){
    if(c[i]==1) seg.update(x[i],(1<<y[i]));
    else cout << __builtin_popcount(seg.query(x[i],y[i],0,0,seg.size)) << "\n";
  }
  
  return 0;
}