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 ll long long int
#define rep(i,n) for( int i = 0; i < n; i++ )
#define rrep(i,n) for( int i = n; i >= 0; i-- )
#define REP(i,s,t) for( int i = s; i <= t; i++ )
#define RREP(i,s,t) for( int i = s; i >= t; i-- )
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define INF 2000000000
#define mod 1000000007
#define INF2 1000000000000000000
class BIT {
    private:
        int size_;
        vector<int> bit;
    public:
        BIT() {
            size_ = 0;
            bit = vector<int>(0);
        }
        BIT(int size__) {
            size_ = size__;
            bit = vector<int>(1000010, 0);
        }
        int sum(int i) {
            int ret = 0;
            while(i > 0) {
                ret += bit[i];
                i -= i & (-i);
            }
            return ret;
        }
        void add(int i, int x) {
            while(i <= size_) {
                bit[i] += x;
                i += i & (-i);
            }
        }
};

int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; cin >> N;
    string S; cin >> S;
    BIT Bit[26];
    rep(i, 26) Bit[i] = BIT(N);
    rep(i, N) {
        Bit[S[i] - 'a'].add(i + 1, 1);
    }
    int M; cin >> M;
    while(M--) {
        int t;
        cin >> t;
        if (t == 2) {
            int l, r;
            cin >> l >> r;
            int cnt = 0;
            rep(i, 26) {
                if (Bit[i].sum(r) - Bit[i].sum(l - 1) > 0) cnt++;
            }
            cout << cnt << endl;
        } else {
            int k; char c;
            cin >> k >> c;
            Bit[S[k - 1] - 'a'].add(k, -1);
            Bit[c - 'a'].add(k, 1);
        }
    }


    return 0;
}