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 <bits/stdc++.h>

using namespace std;

#define pb push_back
#define NL endl
#define pi 2.0*acos(0.0)
#define mod 1e9+7
#define FOR(i,j,n)    for(ll i=j;i<n;i++)
#define RFOR(i,j,n)    for(ll i=j;i>=n;i--)
#define fastIO ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

typedef long long int ll;
typedef unsigned long long ull;
typedef long double ld;

set<ll>cset[30];

int main(){
    ll n,q,a,b,c,pos,ans = 0;
    string s;
    char d,curr;
    cin>>n>>s>>q;
    FOR(i,0,n){
        cset[s[i] - 'a'].insert(i+1);
    }
    FOR(i,0,26){
        cset[i].insert(1e18);
    }
    while(q--){
        cin>>a;
        if(a == 1){
            cin>>b>>d;
            curr = s[b-1];
            cset[curr - 'a'].erase(b);
            cset[d - 'a'].insert(b);
            s[b - 1] = d;
        }
        else{
            cin>>b>>c;
            ans = 0;
            FOR(i,0,26){
                if(cset[i].size()){
                    auto lower = lower_bound(cset[i].begin(),cset[i].end(),b);
                    if(*lower == 1e18){}
                    else{
                        ll val = *lower;
                        if(val >= b && val <= c) ans++;
                    }
                }
            }
            cout<<ans<<endl;
        }
    }
}