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: Accepted)

#include<bits/stdc++.h>
#define MX 500000
using namespace std;
set<int>tr[4*MX+2];
int a[MX+2];
void merge(int pos)
{
    if(tr[2*pos].size() < tr[2*pos+1].size())
    {
        tr[pos] = tr[2*pos+1];
        for(auto u:tr[2*pos])
            tr[pos].insert(u);
    }
    else
    {
        tr[pos] = tr[2*pos];
        for(auto u:tr[2*pos+1])
            tr[pos].insert(u);
    }
}
void build(int pos, int st, int ed)
{
    if(st==ed)
    {
        tr[pos].insert(a[st]);
        return;
    }
    int mid = st+((ed-st)>>1);
    build(2*pos,st,mid);
    build(2*pos+1,mid+1,ed);
    merge(pos);
}
void update(int pos,int st, int ed, int x, int val)
{
    if(st==ed && st==x)
    {
        tr[pos].clear();
        tr[pos].insert(val);
        return;
    }
    int mid = st+((ed-st)>>1);
    if(x<=mid)update(2*pos,st,mid,x,val);
    else update(2*pos+1,mid+1,ed,x,val);
    merge(pos);
}
set<int>query(int pos, int st, int ed, int l, int r)
{
    if(st>r||ed<l)return {};
    if(st>=l && ed<=r)
    {
        return tr[pos];
    }
    int mid = st+((ed-st)>>1);
    set<int>st1 = query(2*pos,st,mid,l,r);
    set<int>st2 = query(2*pos+1,mid+1,ed,l,r);
    if(st1.size()<st2.size())
    {
        for(auto u:st1)st2.insert(u);
        return st2;
    }
    else
    {
        for(auto u:st2)st1.insert(u);
        return st1;
    }
}

string s,s2;
int main()
{
    int n; cin >> n >> s;
    s = "*" + s;
    for(int i=1; i<=n; i++)a[i] = s[i]-'a'+1;
    build(1,1,n);
    int q; cin >> q;
    while(q--)
    {
        int ty; cin >> ty;
        if(ty==1)
        {
            int p; cin >> p >> s2;
            int c = s2[0]-'a'+1;
            update(1,1,n,p,c);
        }
        else
        {
            int l,r; cin >> l >> r;
            set<int>ans = query(1,1,n,l,r);
            cout << ans.size() << "\n";
        }
    }
    return 0;
}