<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 math import comb, factorial
class Solution:
    def iskpal(self, a, k):
        return list(str(a))==list(reversed(str(a))) and a%k==0
    def countposs(self, d):
        p = sum(d.values())
        b = factorial(p)
        for k in d:
            if d[k]!=0:
                b//=(factorial(d[k]))
        return b
    def countGoodIntegers(self, n: int, k: int) -> int:
        if n==1:
            return sum(1 for i in range(1,10) if i%k==0)
        allkpals = []
        m = n//2
        for x in range(10**(m-1),10**(m)):
            if n%2:
                for y in range(0,10):
                    a = int(str(x)+str(y)+"".join(list(reversed(str(x)))))
                    if a%k==0:
                        allkpals.append(str(a))
            else:
                a = int(str(x)+"".join(list(reversed(str(x)))))
                if a%k==0:
                    allkpals.append(str(a))
        # dall = []
        # for y in range(10**(n-1),10**n):
        #     if self.iskpal(y,k):
        #         dall.append(y)
        # print(allkpals)
        # print(len(allkpals),len(dall))
        ret = 0
        seen = set()
        for x in allkpals:
            u = "".join(sorted(list(x)))
            if u not in seen:
                seen.add(u)
            
                d = Counter(x)

                if '0' not in d:
                    ret += self.countposs(d)
                    #print(x,self.countposs(d))
                else:
                    b = 0
                    for k in d:
                        if k!='0':
                            y = d[k]
                            d[k]-=1
                            b += self.countposs(d)
                            d[k] = y
                    ret += b
                        
        return ret
        
        
            
        
24088