<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:
        return sum(count_rearrangements(key) for key in make_keys(n, k))


def make_keys(n, k):
    basis = make_basis(n)
    keys = set()
    for rest_coefs in itertools.product(range(10), repeat=len(basis)-1):
        base = sum(coef * vector for coef, vector in zip(rest_coefs, basis[1:]))
        for main_coef in get_main_coefs(basis[0] % k, base % k, k):
            combo = basis[0] * main_coef + base
            keys.add("".join(sorted(str(combo))))
    return keys


def make_basis(n):
    lo = 1
    hi = 10 ** (n-1)
    basis = []
    while hi >= lo:
        if hi == lo:
            basis.append(hi)
        else:
            basis.append(hi + lo)
        hi //= 10
        lo *= 10
    return basis

@cache
def get_main_coefs(M, B, k):
    return [coef for coef in range(1, 10) if (coef * M + B) % k == 0]


@cache
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
    
18763