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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.Scanner;

class BinaryIndexTree {
    final int n;
    final long[] bit;
    public BinaryIndexTree(int n) {
        this.n = n;
        int size = Integer.highestOneBit(n) << 1;
        bit = new long[size + 1];
    }
    long sum(int i) {
        long s = 0;
        while (i > 0) {
            s += bit[i];
            i -= i & -i;
        }
        return s;
    }
    void add (int i, long x) {
        while (i <= n) {
            bit[i] += x;
            i += i & -i;
        }
    }
}
public class Main {
    public static void main(String [] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        String s = scanner.next();
        BinaryIndexTree[] binaryIndexTrees = new BinaryIndexTree[26];
        for (int i = 0; i < 26; ++i) {
            binaryIndexTrees[i] = new BinaryIndexTree(n);
        }
        for (int i = 0; i < n; ++i) {
            int current = s.charAt(i) - 'a';
            binaryIndexTrees[current].add(i + 1, 1);
        }
        int query = scanner.nextInt();
        for (int i = 0; i < query; ++i) {
            int type = scanner.nextInt();

            if (type == 1) {
                int a = scanner.nextInt();
                int newChar = scanner.next().charAt(0) - 'a';
                int oldChar = s.charAt(a - 1) - 'a';
                binaryIndexTrees[oldChar].add(a, -1);
                binaryIndexTrees[newChar].add(a, 1);
            } else if (type == 2) {
                int l = scanner.nextInt();
                int r = scanner.nextInt();
                int diff = 0;
                for (int j = 0; j < 26; ++j) {
                    long rSum = binaryIndexTrees[j].sum(r);
                    long lSum = binaryIndexTrees[j].sum(l - 1);
                    if (rSum > lSum) {
                        ++diff;
                    }
                }
                System.out.println(diff);
            }
        }
    }
}