<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:
    table = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800]

    def factorial(self, n):
        return self.table[n]

    def createHash(self, num):
        buf = [0]*10
        for n in num:
            buf[ord(n)-48] += 1
        return tuple(buf)

    def findPermutations(self, phash):
        N = sum(phash)
        zeros = phash[0]
        numerator = (N-zeros) * self.factorial(N-1)
        denominator = 1
        for pcount in phash:
            denominator *= self.factorial(pcount)
        return numerator//denominator

    def countGoodIntegers(self, n: int, k: int) -> int:
        isOdd = n&1
        if isOdd:
            palindromes = self.generateOddPalindromes(n, k)
        else:
            palindromes = self.generateEvenPalindromes(n, k)
        hashes = set([self.createHash(pal) for pal in palindromes])
        return sum([self.findPermutations(phash) for phash in hashes])
    
    def generateOddPalindromes(self, n: int, k: int) -> List[str]:
        if n == 1:
            return [str(num) for num in range(0, 10, k)]
        seg = n//2
        first = 10**(seg-1)
        last = first*10
        opt = []
        for pre_num in range(first, last):
            pre_str = str(pre_num)
            suf_str = pre_str[-1::-1]
            for i in range(10):
                pal_str = pre_str + str(i) + suf_str
                pal_int = int(pal_str)
                if pal_int % k == 0:
                    opt.append(pal_str)
        return opt  

    def generateEvenPalindromes(self, n: int, k: int) -> List[str]:
        seg = n//2
        first = 10**(seg-1)
        last = first*10
        opt = []
        for pre_num in range(first, last):
            pre_str = str(pre_num)
            pal_str = pre_str + pre_str[-1::-1]
            pal_int = int(pal_str)
            if pal_int % k == 0:
                opt.append(pal_str)
        return opt

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