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 rng(i,a,b) for(int i=int(a);i<int(b);i++)
typedef long long ll;

int n;
vector<int> a;
vector< bitset<30> > streea;

void builda(int node, int start, int end){
    if(start == end){
        streea[node].set(a[start]);
    }
    else{
        int mid = (start + end) / 2;
        builda(2*node, start, mid);
        builda(2*node+1, mid+1, end);
        streea[node] = streea[2*node] | streea[2*node+1];
    }
}


void update(int node, int start, int end, int idx, int val){
    if(start == end){
        streea[node].reset(a[idx]);
        a[idx] = val;
        streea[node].set(val); 
    }
    else{
        int mid = (start + end) / 2;
            if(start <= idx and idx <= mid){
                update(2*node, start, mid, idx, val);
            }
            else{
                update(2*node+1, mid+1, end, idx, val);
            }
            streea[node] = streea[2*node] | streea[2*node+1];
    }
}

bitset<30> querya(int node, int start, int end, int l, int r){
    if(r < start || end < l){
        return 0;
    }
    
    if(l <= start && end <= r){
        return streea[node];
    }
    
    int mid = (start + end) / 2;
    return querya(2*node, start, mid, l, r) | querya(2*node+1, mid+1, end, l, r);
    
}




int main() {
    ios::sync_with_stdio(false); cin.tie(NULL);
    cin >> n;
    string s; cin >> s;
    int q; cin >>q;
    
    a.resize(n+1,0);
    streea.resize(4*n+4,0);
    
    rng(i, 0, n){
		a[i+1] = (s[i]-'a') + 1;
	}
	builda(1,1,n);
    
    int type, l, r;
    while(q--){
		cin >> type;
		if(type == 1){
			int pos, val;
			char cc;
			cin >> pos >> cc;
			pos--;
			val = (cc-'a') + 1;
			if(s[pos] != cc){
				update(1,1,n, pos+1, val);
			}
		}
		else if( type == 2){
			cin >> l >> r;
			bitset<30> res = querya(1,1,n,l,r);
			cout<<min(26, int(res.count()))<<endl;
		}
	}
    
    
	return 0;
}