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>
using namespace std;

struct data {
    int let[26];
    data() {
        memset(let, 0, sizeof let);
    }
};

const int N = 500005;
data seg[4 * N];
string s;
int n;

void build(int node, int l, int r)
{
    if (l == r) {
        seg[node].let[s[l] - 'a']++;
        return ;
    }
    int mid, left, right;
    mid = l + (r - l) / 2;
    left = 2 * node;
    right = left + 1;
    build(left, l, mid);
    build(right, mid + 1, r);
    for (int i = 0; i < 26; i++) {
        seg[node].let[i] = seg[left].let[i] + seg[right].let[i];
    }
}

void update(int node, int l, int r, int i, char ch)
{
    if (l == r) {
        seg[node].let[s[l] - 'a']--;
        s[l] = ch;
        seg[node].let[s[l] - 'a']++;
        return ;
    }
    int mid, left, right;
    mid = l + (r - l) / 2;
    left = 2 * node;
    right = left + 1;
    if (i <= mid) update(left, l, mid, i, ch);
    else          update(right, mid + 1, r, i, ch);
    for (int i = 0; i < 26; i++) {
        seg[node].let[i] = seg[left].let[i] + seg[right].let[i];
    }
}

data query(int node, int l, int r, int i, int j)
{
    //cout << l << " " << r << '\n';
    data res;
    if (i > r || j < l) {
        return res;
    }
    if (i <= l && r <= j) {
        return seg[node];
    }
    int mid, left, right;
    mid = l + (r - l) / 2;
    left = 2 * node;
    right = left + 1;
    data r1 = query(left, l, mid, i, j);
    data r2 = query(right, mid + 1, r, i, j);
    for (int i = 0; i < 26; i++) {
        res.let[i] = r1.let[i] + r2.let[i];
    }
    return res;
}

int main( )
{
    cin >> n;
    cin >> s;
    build(1, 0, n - 1);
    int q, op, pos, l, r;
    char ch;
    cin >> q;
    while (q--) {
        cin >> op;
        if (op == 1) {
            cin >> pos >> ch;
            pos--;
            update(1, 0, n - 1, pos, ch);
        }
        else {
            cin >> l >> r;
            l--;
            r--;
            data res = query(1, 0, n - 1, l, r);
            int c = 0;
            for (int i = 0; i < 26; i++) {
                if (res.let[i] != 0) {
                    c++;
                }
            }
            cout << c << "\n";
        }
    }
    return 0;
}