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)

/// Containers Start
#include <algorithm>
#include <bitset>
#include <cstdlib>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
/// C Header Files
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <math.h>
#include <stdio.h>
/// Containers End

using namespace std;

/// Math Start
#define PI acos(-1.0)
#define Pi 3.141592653589793
#define EPS (1e-7)
#define INF (0x3f3f3f3f)
// #define INF 9e+18

/// Math End

/// Extra Start
#define nn '\n'
#define pb push_back
#define ull unsigned long long
#define ll long long
#define MOD 1000000007
#define sz(a) int((a).size())
#define space " "
#define All(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
/// Extra End

#define rep(i, n) for (int i = 0; i < n; i++)
#define Cini(a)                                                                \
    int a;                                                                     \
    cin >> a;
#define Cinii(a, b)                                                            \
    int a, b;                                                                  \
    cin >> a >> b;
#define Ciniii(a, b, c)                                                        \
    int a, b, c;                                                               \
    cin >> a >> b >> c;
#define Cins(s)                                                                \
    string s;                                                                  \
    cin >> s;
#define Cinss(s1, s2)                                                          \
    string s1, s2;                                                             \
    cin >> s1 >> s2;
#define Cinc(c)                                                                \
    char c;                                                                    \
    cin >> c;
/// Functions End

/// Debug Start
#define deb(x) cout << #x << ": " << x << endl
#define deb2(x, y) cout << #x << ": " << x << '\t' << #y << ": " << y << endl;
#define deb3(x, y, z)                                                          \
    cout << #x << ": " << x << '\t' << #y << ": " << y << '\t' << #z << ": "   \
         << z << endl;
/// Debug End

// int dx[] = {-1, 0, 1, 0};
// int dy[] = {0, -1, 0, 1};
int dx[] = {-1, 0, 1};
int dy[] = {-1, 0, 1};

/**>>>>>>>>>>>>>>>>>>> END <<<<<<<<<<<<<<<<<<**/

/// template

#define int long long

int N;

void bitAdd(int n,vector<int>& v,int plus){
    for(int i=n;i<=N;i+= i&-i){
        v[i]+= plus;
    }
}

int bitSum(int n, vector<int>& v){
    int ret = 0;
    for(int i=n;i>0;i-=i&(-i)){
        ret += v[i];
    }
    return ret;
}

signed main(void) {
    cin.sync_with_stdio(false);
    cin.tie(0);
    cin >> N;
    Cins(S);
    vector<vector<int> > v2(26,vector<int>(N+1,0LL));
    rep(i,N){
        bitAdd(i+1,v2[(S[i]-'a')],1);
    }
    Cini(Q);
    vector<int> ansV;
    rep(q,Q){
        int type;
        cin >> type;
        if(type==1){
            int i;
            char c;
            cin >> i >> c;
            bitAdd(i,v2[(S[i-1]-'a')],-1);
            bitAdd(i,v2[(c-'a')],1);
        }
        if(type==2){
            int l,r;
            cin >> l >> r;
            int ans = 0;
            rep(i,26) ans += bitSum(r,v2[i])-bitSum(l-1,v2[i]);
            ansV.push_back(ans);
        }
    }
    for(auto i:ansV) cout << i << nn;
    return 0;
}