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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

int conp(char *s,int l,int r){
	int check[500000]={0};
	int count=0,num=r-l+1;
	
	for(int i=l-1;i<(r-l+1)/2;i++){
		count=0;
		if(check[i]==0){
			for(int j=l-1;j<r;j++){
				if(check[j]==0){
					if(s[i]==s[j]){
						count++;
						check[j]=1;
					}
				}
			}
			num=num-count+1;
		}
		check[i]=1;
	}
	
	return num;
}

int main(void){
	int n,q;
	char s[500000];
	scanf("%d",&n);
	scanf("%s",s);
	scanf("%d",&q);
	
	int que,k,l,r;
	char c;
	char c1;
	for(int i=0;i<q;i++){
		scanf("%d",&que);
		switch(que){
		  case 1:
			scanf("%d",&k);
			scanf("%c",&c1);
			scanf("%c",&c);
			if(s[k-1]!=c){
				s[k-1]=c;
			}
			break;
			
		  case 2:
			scanf("%d",&l);
			scanf("%d",&r);
			printf("%d\n",conp(s,l,r));
			break;
		}
	}
}