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 <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <numeric>
#include <string>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <bitset>

typedef long long ll;
typedef unsigned long long ull;
using namespace std;

#define FALSE printf("false\n");
#define TRUE printf("true\n");
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i = 0;i < ((int)(n));++i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define test(a) cout<<"line:"<<__LINE__<<"["<<(#a)<<": "<<(a)<<"]"<<endl

const int INF = 1e9+7;
const ll INFL = 9*1e18;

// const int dx[4] = { 1, 0, -1, 0 };
// const int dy[4] = { 0, 1, 0, -1 };

// const int dx[8] = {1, 1, 0,-1,-1,-1, 0, 1};
// const int dy[8] = {0, 1, 1, 1, 0,-1,-1,-1};

#define SUM(vec) accumulate(all(vec),0LL) // 配列の中身を全部足す
template<typename T>void print(T&& x){cout<<x<<endl;}// 改行付き出力
ll inline digit(ll num){int tmp = 0;while(num){tmp++;num/=10;}return tmp;} // 桁数
template<typename T>inline T digitSum(T num){T sum = 0;while(num){sum+=num%10;num/=10;}return sum;} // 各桁の和
template<typename T>inline T gcd(T a,T b){if(b == 0)return a;return gcd(b,a%b);} // 最大公約数
template<typename T>inline T lcm(T a, T b){T g = gcd(a,b);return a/g*b;} // 最小公倍数
template<typename T>inline T power(T a,T b){T tmp=1;rep(i,b){tmp*=a;}return tmp;} // 累乗
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
int ctoi(char c){return c-'0';}

// TODO: write
int main() {
  int n;cin>>n;
  string s;cin>>s;
  int q;cin>>q;
  
  vector<vector<int>> c(n,vector<int>(26,0));
  vector<vector<int>> dif(n,vector<int>(26,0));
  rep(i,n)c[i][s[i]-'a']++;

  rep(i,q){
    int type;cin>>type;

    if(type == 1){
      int a,b;cin>>a;
      char tmp;cin>>tmp;b = tmp - 'a';
      s[a-1] = tmp;
    }else{
      int a,b;cin>>a>>b;
      vector<bool> tmp(26,0);
      for(int j = a-1;j < b;++j){
        tmp[s[j]-'a'] = true;
      }
      int ans = 0;
      rep(j,26)ans += tmp[j];
      print(ans);
    }
  }

  return 0;
}