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;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<LL,LL> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
const LL mod=1000000007;
const LL LINF=1LL<<62;
const int INF=1<<30;
int dx[]={1,0,-1,0,1,-1,1,-1};
int dy[]={0,1,0,-1,1,-1,-1,1};

const int M_N=600000;

struct BIT{
    //1-indexed
    int n;
    vector<int> bit;
    void init(int k){
        n = k;
        bit.assign(M_N,0);
    }
    //iまでの総和を返す
    LL sum(int i){
        LL s=0;
        while(i>0){
            s+=bit[i];
            i=i&(i-1);
        }
        return s;
    }
    //iにx加算する
    void add(int i,LL x){
        while(i<=n){
            bit[i]+=x;
            i+=i&-i;
        }
    }
};


int main(){
    int n;cin >> n;
    string s;cin >> s;
    BIT bit[26];
    for (int i = 0; i < 26; i++) {
        bit[i].init(n);
    }
    for (int i = 0; i < s.length(); i++) {
        for (int j = 0; j < 26; j++) {
            if(s[i] - 'a' == j) bit[j].add(i+1, 1);
        }
    }
    int q;cin >> q;
    while(q--){
        int t;cin >> t;
        if(t==1){
            int l;
            char c;cin >> l >> c;
            int r = c - 'a';
            int a = bit[r].sum(l) - bit[r].sum(l-1);
            if(a==0) bit[r].add(l,1);
            for (int i = 0; i < 26; i++) {
                if(i==r) continue;
                a = bit[i].sum(l) - bit[i].sum(l-1);
                if(a==1) bit[i].add(l,-1);
            }
        }
        else{
            int l,r;cin >> l >> r;
            int ans = 0;
            for (int i = 0; i < 26; i++) {
                if(bit[i].sum(r) - bit[i].sum(l-1) >= 1) ans++;
            }
            cout << ans << endl;
        }
    }
    return 0;
}