<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:
        
        if n == 1:
            return 9 // k
        
        d = {}
        rem = []
        
        for i in range(n):
            rem.append((10**i) % k)
        
        for i in range(1, 10):
            d[(i,)] = ((rem[0] + rem[-1]) * i) % k
                        
            
        for i in range(1, n//2):
            
            tmp = {}
            for j in range(10):
                crem = ((rem[i] + rem[-1-i]) * j) % k
                for k2,v in d.items():
                    nk = k2 + (j,)
                    nv = (v + crem) % k
                    tmp[nk] = nv
            d = tmp
            
            
        
        ans = 0
        
        ### no mid
        if n % 2 == 0:
            res = set()
            
            for k2,v in d.items():
                if v == 0:
                    res.add(tuple(sorted(k2)))
            
            tot = n
            
            for k2 in res:
                d2 = Counter(k2)
                cnum = 1
                tot2 = tot
                if 0 in d2:
                    cnum *= comb(tot2 - 1, d2[0]*2)
                    tot2 -= d2[0]*2
                
                for k3,v3 in d2.items():
                    if k3 == 0:
                        continue
                    cnum *= comb(tot2, v3*2)
                    tot2 -= v3*2
                ans += cnum
                
            return ans
    
    
        #######
        ### mid
                    
        for j in range(10):
            
            res = set()
            crem = (rem[n//2] * j) % k
            
            for k2,v in d.items():
                if (v + crem) % k == 0:
                    res.add(tuple(sorted(k2)))
            
            tot = n
            
            for k2 in res:
                d2 = Counter(k2)
                
                for k3 in d2:
                    d2[k3] *= 2
                    
                if j not in d2:
                    d2[j] = 1
                else:
                    d2[j] += 1
                
                
                cnum = 1
                tot2 = tot
                if 0 in d2:
                    cnum *= comb(tot2 - 1, d2[0])
                    tot2 -= d2[0]
                
                for k3,v3 in d2.items():
                    if k3 == 0:
                        continue
                    cnum *= comb(tot2, v3)
                    tot2 -= v3
                ans += cnum
                
        return ans
            
                    
                
        
        
32608