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)

// E - Simple String Queries
// ordered set version
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)

int main(){
	int N,Q; string S; cin>>N>>S>>Q;
	vector<set<int>> abc(26);
	rep(i,N) abc[S[i]-'a'].insert(i);
	rep(i,Q){
		int t; cin>>t;
		if(t-1){//type2 display
			int l,r;cin>>l>>r; --l; --r;
			int ans = 0;
			rep(i,26){
				auto it = abc[i].lower_bound(l);
				if(it != abc[i].end() && *it <= r) ans++;
			}
			printf("%d\n",ans);
		}
		else{//type1 update
			int i; char c; cin>>i>>c; --i;
			if(S[i] == c) continue;
			abc[S[i]-'a'].erase(i);
			S[i] = c;
			abc[S[i]-'a'].insert(i);
		}
	}
}