<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:

        ## 1 % k, 10 % k, 100 % k...
        pow10 = [1] * n
        for i in range(1, n):
            pow10[i] = (pow10[i-1] * 10) % k
        # print(pow10) 

        trial = [9] * n
        palins = set()

        ## whether we can solve filling in trial[i:-i]
        ## seek remainder r
        # @cache
        def dp(i, r):
            if (i == (n + 1) // 2): ## reached the end
                # print(i,r)
                if (r == 0):
                    # print(trial)
                    palins.add(
                        tuple((k,v) for k,v in sorted(Counter(trial).items()))
                    )
                return
            
            userange = range(9,0,-1) if i==0 else range(9,-1,-1)
            for d in userange:
                trial[i] = trial[~i] = d ## set digits i-away from ends --> d
                coeff = sum(pow10[j] for j in {i, n-1-i})
                r2 = (r + coeff * d) % k
                # print(i,r, coeff, r2, trial)
                dp(i + 1, r2)
            trial[i] = trial[~i] = 9


        dp(0, 0)
        # print(palins)

        ans = 0
        ## all digit counts of k-palindromes length n
        for p in palins:
            combs = 1
            ct0 = 0
            for k,v in p:
                if k==0: 
                    ct0 = v
                else: 
                    combs *= factorial(v)
            ans += factorial(n) / (combs*factorial(ct0))
            if ct0 > 0: 
                ans -= factorial(n-1) / (combs*factorial(ct0-1))
        
        return int(ans)

        ## all unique digit counts: how many ways to rearrange (no leading 0s)?

31543