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 "bits/stdc++.h"

using namespace std;

using ll = long long;

template<typename T> using V = std::vector<T>;
using Vi = V<int>;
using VVi = V<V<int>>;
using Vl = V<ll>;
using VVl = V<V<ll>>;
using Vs = V<string>;

template<typename T1, typename T2> using P = std::pair<T1, T2>;
using Pii = P<int, int>;
using Pll = P<ll, ll>;
using Pdd = P<double, double>;

template<typename T1, typename T2> using M = std::map<T1, T2>;
using Mii = M<int, int>;
using Msi = M<string, int>;

#define REP(i,n) for(int i=0;i<(int)(n);++i)
#define REP2(i,s,e) for(int i=(int)(s);i<(int)(e);++i)
#define RREP(i,s,e) for(int i=(int)(s);i>=(int)(e);--i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin(),ie=(c).end();i!=ie;++i)
#define ALL(c) (c).begin(), (c).end()

const double PI = acos(-1);
const ll MOD = 1e9+7;

int n, Q;
string s;

int main() {
    cin >> n >> s >> Q;
    V<set<int>> v(26, set<int>());
    REP(i, n) {
        v[s[i] - 'a'].insert(i);
    }

    REP(q, Q) {
        int mode; cin >> mode;
        if (mode == 1) {
            int i; char c; cin >> i >> c; i--;
            if (s[i] == c) continue;
            v[s[i] - 'a'].erase(i);
            v[c - 'a'].insert(i);
            s[i] = c;
        } else if (mode == 2) {
            int l, r; cin >> l >> r; l--; r--;
            int cnt = 0;
            REP(i, 26) {
                set<int> u = v[i];
                auto iter = lower_bound(ALL(u), l);
                if (iter != u.end() && *iter <= r) cnt++;
            }
            std::cout << cnt << '\n';
        }
    }

    return 0;
}