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>

using namespace std;

#define lli long long int
#define REP(i,s,n) for(int i=s;i<n;i++)
#define MOD 1000000007
#define NUM 2520
#define DEBUG 0
#define mp(a,b) make_pair(a,b)
#define SORT(V) sort(V.begin(),V.end())
#define PI (3.141592653589794)


signed main(){

	lli n;
	string s;
	cin>>n>>s;

    set<lli> sets[26];
    REP(i,0,n)sets[s.at(i)-'a'].insert(i);

	lli q;
	cin>>q;

	REP(i,0,q){
		lli k;
		cin>>k;
		if(k==1){
			lli index;
			char a;
			cin>>index>>a;
            index--;
            sets[s.at(index)-'a'].erase(index);
            s[index] = a;
            sets[s.at(index)-'a'].insert(index);
		}
		else{
			lli x,y;
			cin>>x>>y;
			x--,y--;
            lli cnt=0;
            REP(j,0,26){
                set<lli> &nowS = sets[j];
                if(nowS.size()==0)continue;
                auto itr = nowS.lower_bound(x);
                if(DEBUG)cout<<"j="<<j<<" ITR="<<*itr<<endl;
                if(x<=*itr && *itr<=y)cnt++;
            }
            cout<<cnt<<endl;
		}
	}

	return 0;
}