<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>

def get_first():
    return [*map(lambda x: x[0], [*combinations_with_replacement('123456789', 1)])]
def get_digits():
    return [*map(lambda x: x[0], [*combinations_with_replacement('0123456789', 1)])]
def perm(n, r):
    return factorial(n) // factorial(n-r)
def comb(n, r):
    return factorial(n) // (factorial(r)*factorial(n-r))
def pals(n: int) -> List[int]:
    if n == 1:
        return get_first()
    if n%2==0:
        m=n//2
        return [*map(lambda x: ''.join(x) + ''.join(x)[::-1], product(get_first(), *[get_digits() for _ in range(m - 1)]))]
    else:
        return [*map(lambda x: [x[:len(x)//2] + d + x[len(x)//2:] for d in get_digits()], pals(n-1))]
class Solution:
    def countGoodIntegers(self, n: int, k: int) -> int:
        p = pals(n)
        if n%2==1:
            p = [y for x in p for y in x]
        c = set(frozenset(Counter(x).items()) for x in p if int(x) % k == 0)
        ans=0
        for fst in c:
            N = 0
            num_zeros = 0
            mult = 1
            # need to handle no leading zeroes
            for item in fst:
                N += item[1]
                rs = []
                if item[0] == '0':
                    num_zeros += item[1]
                else:
                    mult *= factorial(item[1])
            ans += comb(N-1, num_zeros) * (factorial(N-num_zeros) / (mult))
        #120021
        #102201
        print(ans)
        return int(ans)
27016