<p>You are given two <strong>positive</strong> integers <code>n</code> and <code>k</code>.</p>

<p>An integer <code>x</code> is called <strong>k-palindromic</strong> if:</p>

<ul>
	<li><code>x</code> is a <span data-keyword="palindrome-integer">palindrome</span>.</li>
	<li><code>x</code> is divisible by <code>k</code>.</li>
</ul>

<p>An integer is called <strong>good</strong> if its digits can be <em>rearranged</em> to form a <strong>k-palindromic</strong> integer. For example, for <code>k = 2</code>, 2020 can be rearranged to form the <em>k-palindromic</em> integer 2002, whereas 1010 cannot be rearranged to form a <em>k-palindromic</em> integer.</p>

<p>Return the count of <strong>good</strong> integers containing <code>n</code> digits.</p>

<p><strong>Note</strong> that <em>any</em> integer must <strong>not</strong> have leading zeros, <strong>neither</strong> before <strong>nor</strong> after rearrangement. For example, 1010 <em>cannot</em> be rearranged to form 101.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 3, k = 5</span></p>

<p><strong>Output:</strong> <span class="example-io">27</span></p>

<p><strong>Explanation:</strong></p>

<p><em>Some</em> of the good integers are:</p>

<ul>
	<li>551 because it can be rearranged to form 515.</li>
	<li>525 because it is already k-palindromic.</li>
</ul>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 1, k = 4</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p>The two good integers are 4 and 8.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 5, k = 6</span></p>

<p><strong>Output:</strong> <span class="example-io">2468</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10</code></li>
	<li><code>1 &lt;= k &lt;= 9</code></li>
</ul>

class Solution:
    def countGoodIntegers(self, n: int, k: int) -> int:
        num, seen = ['#'] * n, set()

        def hash_digits(num):
            digits_hash = ''
            count = [0] * 10
            for x in map(ord, num):
                count[x - 48] += 1
            for c in count:
                digits_hash += chr(c + 48) + '#'
            return digits_hash

        def F(num):
            count = [0] * 10
            res = factorial(n)
            for x in map(ord, num):
                count[x - 48] += 1
            denominator = 1
            for digit in range(1, 10):
                if count[digit]:
                    denominator *= factorial(count[digit])
            res //= denominator * factorial(count[0])
            if count[0]:
                denominator *= factorial(count[0] - 1)
                res -= factorial(n - 1) // denominator
            return res

        def solve(idx, first):
            count = 0
            if idx == n // 2:
                if n & 1:
                    for digit in range(int(first), 10):
                        num[idx] = chr(digit + 48)
                        x = int(''.join(num))
                        digits_hash = hash_digits(num)
                        if x % k == 0 and digits_hash not in seen:
                            seen.add(digits_hash)
                            count += F(num)
                else:
                    x = int(''.join(num))
                    digits_hash = hash_digits(num)
                    if x % k == 0 and digits_hash not in seen:
                        seen.add(digits_hash)
                        count += F(num)
                return count
            else:
                for digit in range(int(first), 10):
                    num[idx] = num[-idx - 1] = chr(digit + 48)
                    count += solve(idx + 1, False)
                return count

        return solve(0, True)
25153