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: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0; i < n; i++)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define ll long long
#define MOD 1000000007
#define MAX 100010
#define aout(a) REP(i,sizeof(a)) cout << a[i] << " "; cout << endl;
#define vout(v) REP(i,v,size()) cout << v[i] << " "; cout << endl;
using namespace std;
using Graph = vector<vector<int>>;

int main(){
  int n; cin >> n;
  string s; cin >> s;
  int q; cin >> q;

  REP(i,q){
    int a; cin >> a;
    int ans = 0;
    vector<int> count(26,0);
    if(a == 1){
      int b; cin >> b;
      string c; cin >> c;
      s.erase(b-1,1);
      s.insert(b-1,c);
      //cout << s << endl;
    }
    else{
      int d; cin >> d;
      int e; cin >> e;
      //FOR(j,d-1,e) count[s[j]-'a']++;
      REP(j,26) if(count[j] != 0) ans++;
      cout << ans << endl;
    }
  }
}