<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>

from itertools import permutations
class Solution:
    def countGoodIntegers(self, n: int, k: int) -> int:
        ans = 0
        if n == 1:
            for i in range(1, 10):
                if i % k == 0:
                    ans += 1
            return ans                    
        n2 = n // 2
        visited = set()
        def makehist(s):
            ans = [0] * 10
            for c in s:
                ans[int(c)] += 1
            return tuple(ans)  
        @lru_cache(None)
        def hist2ans(hist, i):
            if i == n:
                return 1
            ans = 0
            vhist = list(hist)      
            for j in range(10):
                if i == 0 and j == 0:
                    continue
                if vhist[j] == 0:
                    continue
                vhist[j] -= 1
                ans += hist2ans(tuple(vhist), i + 1)
                vhist[j] += 1
            return ans                
        def sub(s):
            n = len(s)
            s2 = s[:(n // 2)]            
            mid = s[(n // 2):(- (n // 2))]
            s3 = s2 + mid + s2[::-1]
            if int(s3) % k != 0:
                return 0
            hist = makehist(s)
            #print(f'hist({s})={hist}')
            if hist in visited:
                return 0
            visited.add(hist)
            return hist2ans(hist, 0)
        for i in range(10 ** (n2 - 1), 10 ** n2):
            if n % 2 == 1:
                for j in range(10):
                    ans += sub(str(i) + str(j) + str(i)[::-1])
            else:
                ans += sub(str(i) + str(i)[::-1])
        return ans                
                
            
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