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 PHP (7.4.4) (judge verdict: Time Limit Exceeded)

<?php
list($n) = ints();
list($s) = strs();
for ($i = 0; $i < $n; ++$i) $a[] = 1 << (ord($s[$i]) - 97);
$st = new SegmentTree($a);
list($q) = ints();
for ($i = 0; $i < $q; ++$i) {
    list($type, $a, $b) = strs();
    if ($type === '1') {
        $st->set($a - 1, 1 << (ord($b) - 97));
    } else {
        $output[] = popcount($st->get($a - 1, $b));
    }
}
echo implode(PHP_EOL, $output);
function strs()
{
    return explode(' ', trim(fgets(STDIN)));
}
function ints()
{
    return array_map('intval', explode(' ', trim(fgets(STDIN))));
}
function popcount($x)
{
    $res = 0;
    while ($x > 0) {
        $x &= $x - 1;
        ++$res;
    }
    return $res;
}
// セグメントツリー
// 更新と取得を O(logN) で行う
class SegmentTree
{
    private $n;
    private $node;
    const UNIT = 0; // 単位元 問題に応じて変更する

    public function __construct(array $a)
    {
        $l = count($a);
        $n = 1;
        while ($n < $l) $n *= 2;
        $node = array_fill(0, 2 * $n - 1, self::UNIT);
        for ($i = 0; $i < $l; ++$i) $node[$i + $n - 1] = $a[$i];
        $this->n = $n;
        $this->node = $node;
        $this->updateAll();
    }

    private function updateAll(): void
    {
        $n = $this->n;
        $node = $this->node;
        for ($i = $n - 2; $i >= 0; --$i)
            $node[$i] = op($node[2 * $i + 1], $node[2 * $i + 2]);
        $this->node = $node;
    }

    public function set(int $x, int $v): void
    {
        $n = $this->n;
        $node = $this->node;
        $x += $n - 1;
        $node[$x] = $v;
        while ($x > 0) {
            $x = intdiv($x - 1, 2);
            $node[$x] = op($node[2 * $x + 1], $node[2 * $x + 2]);
        }
        $this->node = $node;
    }

    // 区間 [a, b)
    public function get(int $a, int $b, int $k = 0, int $l = 0, int $r = -1): int
    {
        $n = $this->n;
        $node = $this->node;
        if ($r < 0) $r = $n;
        // 範囲外
        if ($a >= $r || $b <= $l) return self::UNIT;
        // 完全に含む
        if ($a <= $l && $b >= $r) return $node[$k];
        // 一部含む
        return op(
            $this->get($a, $b, 2 * $k + 1, $l, intdiv($l + $r, 2)),
            $this->get($a, $b, 2 * $k + 2, intdiv($l + $r, 2), $r)
        );
    }
}
// 二項演算 問題に応じて変更する
function op(int $a, int $b): int
{
    return $a | $b;
}