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;
typedef long long ll;
typedef vector<long long> vll;
typedef vector<pair<int, int>> vpi;
typedef vector<pair<long long, long long>> vpl;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
#define INF (int)(1e9)
#define MAXX 1.1529215e+18
#define inf 999999
#define EPS (1e-7)
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i, n) for (int i = 1; i <= (int)(n); i++)
#define FOR(i, k, n) for (int i = (k); i < (int)(n); i++)
#define ALL(a) a.begin(), a.end()
#define RALL(a) a.begin(), a.end(), greater<int>()
#define ROT(a) a.begin(), a.begin() + 1, a.end()
#define RROT(a) a.begin(), a.end() - 1, a.end()
#define PB push_back
#define MP make_pair
#define PI acos(-1.0)
#define sz(a) a.size()
const ll MOD = 1e9 + 7;
const int MAX = 10000000;
template <typename T1, typename T2>
inline bool chmin(T1& a, T2 b)
{
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T1, typename T2>
inline bool chmax(T1& a, T2 b)
{
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
T gcd(T a, T b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
//cout << 'H' << endl;
/*--------------------------------------------*/
const int MAX_N = 10000000;

int n;
string s;
bitset<26> dat[MAX_N];

void update(int k, char a);
void init(int n_)
{
    n = 1;
    while (n < n_)
        n *= 2;

    rep(i, n_)
    {
        update(i, s[i]);
    }
}

//k番目の値をaに変更
void update(int k, char a)
{
    k += n - 1;

    dat[k] = 0;
    dat[k].set(a - 'a', true);

    while (k > 0) {
        k = (k - 1) / 2;
        dat[k] = (dat[k * 2 + 1] | dat[k * 2 + 2]);
    }
}

//[a,b)の最小値を求める

bitset<26> query(int a, int b, int k, int l, int r)
{
    if (r <= a || b <= l)
        return bitset<26>(0);
    if (a <= l && b >= r)
        return dat[k];

    bitset<26> vl = query(a, b, 2 * k + 1, l, (l + r) / 2);
    bitset<26> vr = query(a, b, 2 * k + 2, (l + r) / 2, r);

    return (vl | vr);
}

int main()
{
    //cout << fixed << setprecision(10)
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N;
    int Q;
    cin >> N >> s >> Q;

    init(N);


    rep(_, Q)
    {
        int q;
        cin >> q;
        --q;
        if (!q) {
            int i;
            char c;
            cin >> i >> c;
            --i;
            update(i, c);
        } else {
            int l, r;
            cin >> l >> r;
            --l;
            auto ans = query(l, r, 0, 0, n);
            cout << ans.count() << endl;
        }
    }
}