<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:
        @cache
        def factorial(n):
            n = int(n)
            sum = 1
            for i in range(1, n + 1):
                sum *= i
            return int(sum)
        
        def get_good(k_par):
            cnt = Counter(k_par)
            zeros = cnt['0']
            prod = 1
            for d, p in cnt.items():
                prod *= factorial(p)
            ret = (len(k_par) - zeros) * factorial(len(k_par) - 1) // prod
            return ret

        digits = "0123456789"
        ans = 0
        vis = set()
        def dfs(cur, leading_zero):
            nonlocal ans, vis
            if len(cur) == n // 2:
                postfix = cur[::-1]
                if n % 2 == 0:
                    value = int(cur + postfix)
                    strr = str(cur + postfix)
                    if value % k == 0 and ''.join(sorted(strr)) not in vis:
                        vis.add(''.join(sorted(strr)))
                        ans += get_good(strr)
                else:
                    for c in digits:
                        value = int(cur + c + postfix)
                        strr = str(cur + c + postfix)
                        if value % k == 0 and ''.join(sorted(strr)) not in vis:
                            vis.add(''.join(sorted(strr)))
                            ans += get_good(strr)
                return 

            for i in range(10):
                c = digits[i]
                if leading_zero and i == 0:
                    continue
                dfs(cur + c, False)

        dfs("", True)
        return ans
            
23023