<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>

BASIS = [
    [],
    [1],
    [11],
    [101, 10],
    [1001, 110],
    [10001, 1010, 100],
    [100001, 10010, 1100],
    [1000001, 100010, 10100, 1000],
    [10000001, 1000010, 100100, 11000],
    [100000001, 10000010, 1000100, 101000, 10000],
    [1000000001, 100000010, 10000100, 1001000, 110000],
]


class Solution:
    def countGoodIntegers(self, n: int, k: int) -> int:
        return sum(count_rearrangements(key) for key in make_keys(n, k))


def make_keys(n, k):
    basis = BASIS[n]
    keys = set()
    combiner = Combiner(basis)
    for combo in combiner.make_combinations():
        if combo >= basis[0] and combo % k == 0:
            keys.add("".join(sorted(str(combo))))
    return keys


class Combiner:
    def __init__(self, basis):
        self.basis = basis
    
    def make_combinations(self, j=0, prev=0):
        if j == len(self.basis):
            return [prev]
        ans = []
        for coef in range(10):
            leader = coef * self.basis[j]
            ans.extend(self.make_combinations(j=j+1, prev=prev+leader))
        return ans


def count_rearrangements(key):
    counts = collections.Counter(key)
    zeros = counts["0"]
    other_counts = tuple(sorted(v for k, v in counts.items() if k != "0"))
    return cached_count(zeros, other_counts, False)


@cache
def cached_count(zeros, other_counts, allow_leading_zero):
    if (zeros, other_counts) == (0, ()):
        return 1
    ans = 0
    if allow_leading_zero and zeros > 0:
        ans += cached_count(zeros - 1, other_counts, True)
    for j in range(len(other_counts)):
        dec = list(other_counts)
        dec[j] -= 1
        dec = tuple(sorted(x for x in dec if x > 0))
        ans += cached_count(zeros, dec, True)
    return ans
    
22756