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 ff first
#define ss second
#define len(x) (int)(x.size())
#define all(vs) vs.begin(), vs.end()
using ii = pair<int, int>;
using ll = long long;
#define MAX 500010
char st[MAX];
const int neutral = 0;
#define comp(a, b) ((a)+(b))
using vi = vector<int>;

class SegTree{
	vi a; //using vi = vector<int>
	int n;
	public:
	SegTree(){
		int sz = MAX;
		for(n = 1; n<sz; n<<=1 );
		a.assign(n<<1, neutral);
		for(int i=0; i<sz; ++i) a[i+n] = st[i];
		for(int i=n+sz-1; i>1; --i)
			a[i>>1] = comp(a[i>>1], a[i]);
	}
	void update(int i, int x){
		a[i+=n] = x; 
		for( i>>=1; i ; i>>=1)
			a[i] = comp(a[i<<1], a[1+(i<<1)]);
	}
	int rsq(int l, int r){
		int ans = neutral;
		for(l+=n, r+=n+1; l<r; l>>=1, r>>=1){
			if(l&1) ans=comp(ans, a[l++]);
			if(r&1) ans=comp(ans, a[--r]);
		}
		return ans;
	}
};

SegTree bt[26];
int32_t main(){
	int n;
	scanf("%d", &n);
	scanf("%s", st);
	for(int i=0; i<n; ++i)
		bt[st[i]-'a'].update(i, 1);
	int q;
	scanf("%d", &q);
	while(q--){
		int op;
		scanf("%d ", &op);
		if(op==1){
			int i; 
			char c;
			scanf("%d %c", &i, &c);
			--i;
			bt[st[i]-'a'].update(i, 0);
			bt[c-'a'].update(i, 1);
		}
		else{
			int l, r;
			scanf("%d %d", &l, &r);
			--l; --r;
			int ans = 0;
			for(int j=0; j<26; ++j){
				if(bt[j].rsq(l, r)>0){ 
					++ans;
				}
			}
			printf("%d\n", ans);
		}
	}
}