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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#include<cmath>
#define pb push_back
#define ld long double
#define mp make_pair
#define vl vector<ll> 
#define vd vector<double>
#define vld vector<long double>
#define ll long long int
#define pl pair<ll, ll>
#define all(a) a.begin(), a.end()
#define forr(i, n) for(ll i=0; i<n; i++) 
#define forr1(i, n) for(ll i=1; i<=n; i++)
using namespace std;
const ld PI =3.1415926535897923846;
const ll MOD = 1e9+7;
const ll N=998244353;

struct node{
    vl alpha;
    ll ans;
}segtree[1048679];
string str;
node util(node a, node b)
{
    node an;
    ll num=0;
    forr(i, 26)
    {
        ll total=a.alpha[i]+b.alpha[i];
        an.alpha.pb(total);
        if(total)
        num++;
    }
    an.ans=num;
    return an;
}
node Query(ll start, ll end, ll r1, ll r2, ll put)
{
    if(start>=r1&&end<=r2)
    return segtree[put];
    node sex;
    forr(i, 26)
    sex.alpha.pb(0);
    sex.ans=0;
    if(start>r2||end<r1)
    return sex;
    ll mid=start+(end-start)/2;
    return util(Query(start, mid, r1, r2, 2*put+1), Query(mid+1, end, r1, r2, 2*put+2));
}
node constructTree(ll start, ll end, ll put)
{
    if(start==end)
    {
        forr(i, 26)
        segtree[put].alpha.pb(0);
        segtree[put].alpha[(int)(str[start]-'a')]++;
        segtree[put].ans=1;
        return segtree[put];
    }
    ll mid=start+(end-start)/2;
    segtree[put]=util(constructTree(start, mid, 2*put+1), constructTree(mid+1, end, 2*put+2));
    return segtree[put];
}
node Update(ll start, ll end, ll index, ll put, char ch)
{
    if(index>end||index<start)
    return segtree[put];
    if(start==end)
    {
        if(start==index)
        {
            ll oldie=str[index]-'a';
            ll newie=ch-'a';
            str[index]=ch;
            segtree[put].alpha[oldie]--;
            segtree[put].alpha[newie]++;
            ll num=0;
            forr(i, 26)
            {
                if(segtree[put].alpha[i])
                num++;
            }
            segtree[put].ans=num;
            return segtree[put];
        }
        else
        return segtree[put];
    }
    ll mid=start+(end-start)/2;
    segtree[put]=util(Update(start, mid, index, 2*put+1, ch), Update(mid+1, end, index, 2*put+2, ch));
    return segtree[put];
}
void solve()
{
    ll n, q;
    cin>>n;
    cin>>str;
    cin>>q;
    ll x=(ll)ceil((log2(n)));
    x=(ll)pow(2, x);
    constructTree(0, n-1, 0);
    while(q--)
    {
        ll type, q1, q2;
        cin>>type;
        if(type==2)
        {
            cin>>q1>>q2;
            //cout<<q1<<" "<<q2<<endl;
            --q1;
            --q2;
            node lala=Query(0, n-1, q1, q2, 0);
            cout<<lala.ans<<endl;
        }
        else
        {
            char ch;
            cin>>q1>>ch;
            //cout<<q1<<" "<<ch<<endl;
            --q1;
            Update(0, n-1, q1, 0, ch);
        }
    }
    
}
int main()
{
    ll test=1;
    //cin>>test;
    while(test--)
        solve();
}