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>
#include <stdlib.h>

int main(void) {
    int N, Q, *A, B[30], X, i, j;
    int q, I, l, r;
    char S[500005], c;
    scanf("%d%s%d", &N, S, &Q);
    A=malloc(26*(N+1)*sizeof(int));
    for (i=0; i<26; i++) {
        A[i]=0;
        if (S[0]-97==i) A[i]++;
    }
    for (i=1; i<N; i++) {
        for (j=0; j<26; j++) {
            A[26*i+j]=A[26*(i-1)+j];
            if (S[i]-97==j) A[26*i+j]++;
        }
    }
    for (i=0; i<Q; i++) {
        scanf("%d", &q);
        // printf("[%d]\n", q);
        if (q==1) {
            scanf("%d %c", &I, &c);
            // printf("(%d %c)\n", I, c);
            I--;
            if (S[I]==c) continue;
            S[I]=c;
        }
        else {
            scanf("%d%d", &l, &r);
            // printf("[%d %d]\n", l, r);
            l--; r--;
            X=0;
            for (j=0; j<26; j++) {
                if (l!=0) B[j]=A[26*r+j]-A[26*(l-1)+j];
                else B[j]=A[26*r+j];
                if (B[j]!=0) X++;
            }
            printf("%d\n", X);
        }
    }
    // for (i=0; i<26; i++) printf("%d ", A[i]);
    // printf("%d", A[26*(N-1)+25]);
    return 0;
}