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: Time Limit Exceeded)

#include <stdio.h>

int main()
{
    int n;
    char s[500001];

    int q,t,i,l,r;
    char c;


    scanf("%d %s %d",&n,s,&q);

    int j,ans = 0,k,bucket[26] ={0};

    for (j = 0; j < q; j++)
    {
        scanf("%d",&t);

        if(t == 1)
        {
            scanf("%d %c",&i,&c);
            s[i - 1] = c;
        }
        else
        {
            scanf("%d %d",&l,&r);
            for (k = 0; k < 26; k++)
            {
                bucket[k] = 0;
            }
            ans = 0;
            for (k = l - 1; k <= r - 1; k++)
            {
                bucket[s[k] - 'a'] = 1;
            }
            for (k = 0; k < 26; k++)
            {
                ans += bucket[k];
            }
            printf("%d\n",ans);
        }
        
    }
    
}