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: Runtime Error)

#include <bits/stdc++.h>

#define rep(i, a, n)  for(ll i=a; i<n; i++)
#define per(i, a, n)  for(ll i=n-1; i>=a; i--)
#define fill0(n)  setfill('0') << right << setw(n)
#define all(x)  (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define LONGMAX 1e18
#define INTMAX 1000000000

using namespace std;
typedef long long ll;
typedef pair<ll ,ll> P;
const ll mod = 1e9+7;
const long double EPS = 0.0000000001;

int n, q;
string s;
int main(){
	cin >> n;
	cin >> s;
	cin >> q;
	
	// アルファベットの数に対応したsetのvectorを用意する.
	// is[0]のsetにはaが現れる位置を持たせておく。
	// set は insert とサーチが高速;
	vector<set<int>> is(26); 
	for(int i = 0; i < n; i++)is[s[i] - 'a'].insert(i);
	rep(i, 0 ,q){
		int type;
		cin >> type;
		if(type == 1){
			int iq; char cq;
			cin >> iq >> cq;
			iq--;
			is[cq - 'a'].erase(iq);
			s[iq] = cq;
			is[cq - 'a'].insert(iq);
			
		}else {
			int lq, lr;
			cin >> lq >> lr;
			lq--;
			int ans = 0;
			for(int j = 0; j < 26; j++){
				auto itr = is[i].lower_bound(lq);
				if(itr != is[i].end() and *itr < lr )ans++;
			}
			cout << ans << endl;
		}
	}

}