<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:
        mod10 = list(accumulate([1]+[10]*(n-1),lambda x,y:(x*y)%k))
        mods = [(mod10[i] + mod10[~i])%k for i in range(0, n//2)]
        if n % 2: mods.append(mod10[(n-1)//2])
        if k == 1: mods[0] = 1
        while mods[-1]==0: mods.pop()
        evenafter = list(accumulate(mods[::-1], lambda x,y: x&(~y&1), initial=True))[::-1]
        nm = len(mods)
        @cache
        def f(i, m=0):
            res = []
            if i == nm: return [[]] if m==0 else []
            if m&1 and not k&1 and evenafter[i]: return []
            for j in range(10):
                if i==0 and j==0: continue
                res += [[j]+ x for x in f(i+1, (m-j*mods[i])%k)]
            return res
        
        def count(rl):
            def count_full(c):
                return reduce(lambda x,y: x//factorial(y), c.values(), (n-c[0])*factorial(n-1))
            c = Counter(rl); c += Counter(rl[:n//2])
            if len(rl) == (n+1)//2: return count_full(c)
            return sum(count_full(c+Counter([d])) for d in range(10))    
        
        digicomb = set()
        for l in f(0):
            for r in combinations_with_replacement(range(10), max(0, n//2 - nm)):
                t = tuple(sorted(l[:n//2]) + l[-1:]) if nm > n//2 else tuple(sorted(tuple(l) + r))
                digicomb.add(t)

        return sum(count(rl) for rl in digicomb)

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