<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:
        
        def backtrack(cur_ind, length, l1, thing):
            if cur_ind == length:
                l1.append(thing)
                return
            
            if not cur_ind:
                for i in range(1, 10):
                    backtrack(cur_ind + 1, length, l1, thing + str(i))
            else:
                for i in range(10):
                    backtrack(cur_ind + 1, length, l1, thing + str(i))
                    
                    
        def factorial(n):
            res = 1
            
            for i in range(1, n + 1):
                res *= i
                
            return res
        
        def check_zero(number):
            if "0" not in number:
                return 0
            
            l1 = list(number)
            
            l1.remove("0")
            
            fact = factorial(len(l1))
            
            for i in "0123456789":
                fact //= factorial(l1.count(i))
                
            return fact
            
                    
        
        if n == 1:
            if k == 1:
                return 9
            if k == 2:
                return 4
            if k == 3:
                return 3
            if k == 4:
                return 2
            return 1
        
        if n % 2:
            res = 0
            things = []
            backtrack(0, n // 2, things, "")
            seen = set()
            
            for t in things:
                for new_digit in "0123456789":
                    cur_int = int(t + new_digit + t[::-1])
                    string = t + new_digit + t[::-1]
                    
                    if "".join(sorted(string)) in seen:
                        continue

                    if not cur_int % k:
                        added = factorial(len(string))

                        for i in "0123456789":
                            added //= factorial(string.count(i))
                            
                        added -= check_zero(string)
                            
                        seen.add("".join(sorted(string)))

                        res += added
                        
            
            
            return res
        
        
        else:
            res = 0
            things = []
            backtrack(0, n // 2, things, "")
            seen = set()
            
            for t in things:
                string = t + t[::-1]
                cur_int = int(t + t[::-1])
                
                if "".join(sorted(string)) in seen:
                    continue
                
                
                if not cur_int % k:
                    added = factorial(len(string))
                    
                    for i in "0123456789":
                        added //= factorial(string.count(i))
                        
                    added -= check_zero(string)
                        
                    res += added
                
                    
                    seen.add("".join(sorted(string)))
                    
            return res
        
23289