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: Runtime Error)

#include<stdio.h>
int n,B[26][50010]={0};
//BITのadd関数
//i番めの数値にnを足す
void badd(int j,int i,int a){for(;i<=n;i+=i&-i)B[j][i]+=a;}
//BITのsum関数
//1からi番めの数の総和を返す
int bsum(int j,int i){
  int s=0;
  for(;i;i-=i&-i)s+=B[j][i];
  return s;
}
int main(){
  int m,a,b,d,q,i;
  char c,s[500010]={};
  scanf("%d %s",&n,s);//printf("a\n");
  for(i=0;s[i];i++)badd(s[i]-'a',i+1,1);//printf("b\n");
  //printf("%d\n",bsum(0,1));
  scanf("%d",&m);
  while(m--){//printf("%d\n",m);
    scanf("%d",&q);
    if(q==1){
      scanf("%d %c",&a,&c);
      badd(s[a-1]-'a',a,-1);
      s[a-1]=c;
      badd(s[a-1]-'a',a, 1);
    }
    else{
      scanf("%d %d",&a,&b);
      for(i=d=0;i<26;i++){//printf("%d %d\n",bsum(i,b),bsum(i,a-1));
	if(bsum(i,b)!=bsum(i,a-1))d++;
      }
      printf("%d\n",d);
    }
  }
  return 0;
}