<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>

from typing import List
from math import factorial

class Solution:
    MOD = 10 ** 9 + 7

    def countGoodIntegers(self, n: int, k: int) -> int:
        # Precompute factorials and inverse factorials modulo MOD
        max_fact = n
        factorials = [1] * (max_fact + 1)
        inv_factorials = [1] * (max_fact + 1)
        for i in range(1, max_fact + 1):
            factorials[i] = (factorials[i - 1] * i) % self.MOD
        inv_factorials[max_fact] = pow(factorials[max_fact], self.MOD - 2, self.MOD)
        for i in range(max_fact - 1, -1, -1):
            inv_factorials[i] = (inv_factorials[i + 1] * (i + 1)) % self.MOD

        def count_permutations(counts, total_digits):
            total_perms = factorials[total_digits]
            for cnt in counts.values():
                total_perms = (total_perms * inv_factorials[cnt]) % self.MOD
            return total_perms

        def count_permutations_starting_with_non_zero(counts):
            total = 0
            total_digits = sum(counts.values())
            for d in range(1, 10):
                if counts.get(d, 0) > 0:
                    counts[d] -= 1
                    perms = count_permutations(counts, total_digits - 1)
                    total = (total + perms) % self.MOD
                    counts[d] += 1
            return total

        # Generate palindromic numbers of length n
        hlen = (n + 1) // 2  # Half length (including middle digit for odd n)
        start = 10 ** (hlen - 1)
        end = 10 ** hlen
        counts_set = {}
        for first_half in range(start, end):
            first_half_str = str(first_half)
            if first_half_str[0] == '0':
                continue  # Skip numbers with leading zeros
            # Generate full palindrome
            if n % 2 == 0:
                palindrome_str = first_half_str + first_half_str[::-1]
            else:
                palindrome_str = first_half_str + first_half_str[:-1][::-1]
            palindrome_int = int(palindrome_str)
            if palindrome_int % k == 0:
                # Get counts of digits
                counts = [0]*10
                for ch in palindrome_str:
                    d = int(ch)
                    counts[d] += 1
                counts_key = tuple(counts)
                counts_set[counts_key] = counts

        # Now, for each unique counts, compute the permutations once
        total_count = 0
        for counts_list in counts_set.values():
            counts_dict = {i: counts_list[i] for i in range(10) if counts_list[i] > 0}
            perms = count_permutations_starting_with_non_zero(counts_dict)
            total_count = (total_count + perms) % self.MOD

        return total_count

18763