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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<set>
using namespace std;
static const int MAX = 500000;


typedef long long ll;

int n, q;
string s;
set<int> A[26];

void Init(){
    for(int i = 0; i < 26; i++){
        A[i].insert(1000000);
    }
    for(int i = 0; i < n; i++){
        A[s[i] - 'a'].insert(i);
    }
}

void Change(int x, char c){
    x--;
    A[s[x] - 'a'].erase(x);
    A[c - 'a'].insert(x);
    s[x] = c;
}

void Ans(int l, int r){
    l--; r--;
    int ans = 0;
    for(int i = 0; i < 26; i++){
        if(*lower_bound(A[i].begin(), A[i].end(), l) <= r){
            ans++;
        }
    }
    cout << ans << "\n";
}

int main(){
    ios::sync_with_stdio(false);
    cin >> n >> s >> q;

    Init();

    for(int i = 0; i < q; i++){
        int x;
        cin >> x;
        if(x == 1){
            int y;
            char c;
            cin >> y >> c;
            Change(y, c);
        }
        else{
            int l, r;
            cin >> l >> r;
            Ans(l, r);
        }
    }

    return 0;
}