<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:
        ans = [0 for i in range(n)]
        fans = [0,k]
        d = {}
        def int1(digits):
            digits.sort(reverse=True)
            ans = 0
            for i in range(len(digits)):
                ans += digits[i]*(10**(len(digits)-i-1))
            return ans
        def func1(digits):
            if int1(digits[:]) in d:
                return 0
            d[int1(digits[:])] = 1
            digit_count = Counter(digits)
            n = len(digits)
            def factorial_with_repeats(counts):
                num = math.factorial(sum(counts.values()))
                if counts[0]==0:
                    counts[0] = 1
                dem = math.prod(math.factorial(v) for v in counts.values())
                return num // dem
            total_permutations = factorial_with_repeats(digit_count)
            if 0 not in digits:
                return total_permutations
            digit_count[0] -= 1
            total_permutations -= factorial_with_repeats(digit_count)
            return total_permutations
        def func(ans):
            st = ''
            for j in range(len(ans)):
                st += str(ans[j])
            st = int(st)
            if st%(fans[1])==0:
                return func1(ans)
            else:
                return 0
        def rec(i,ans):
            if i>n-i-1:
                fans[0] += func(ans)
                return
            if i==0:
                for j in range(1,10):
                    ans[i] = j
                    ans[n-i-1] = j
                    rec(i+1,ans)
                ans[i] = 0
                ans[n-i-1] = 0
            else:
                for j in range(10):
                    ans[i] = j
                    ans[n-i-1] = j
                    rec(i+1,ans)
                ans[i] = 0
                ans[n-i-1] = 0
        rec(0,ans)
        return fans[0]
20626