<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 count_unique_rearrangements_exclude_leading_zero(digit_string):
            # Count the frequency of each digit
            freq = Counter(digit_string)
            
            # Calculate the factorial of the total length of the string
            total_permutations = factorial(len(digit_string))
            
            # Divide by the factorial of the frequency of each digit
            for count in freq.values():
                total_permutations //= factorial(count)
            
            # If '0' is in the string, calculate permutations where '0' is the leading digit
            if '0' in freq:
                # Subtract the permutations where '0' is the leading digit
                # This is done by considering permutations of the rest of the string
                leading_zero_permutations = factorial(len(digit_string) - 1)
                for digit, count in freq.items():
                    if digit == '0':
                        leading_zero_permutations //= factorial(count - 1)
                    else:
                        leading_zero_permutations //= factorial(count)
                
                total_permutations -= leading_zero_permutations
            
            return total_permutations

        res = [0]
        seen = set()
        def dfs(i, n, curr):
            if i >= n // 2 + n % 2:
                if n % 2:
                    temp, mid_digit = curr[:-1], curr[-1]
                    final = temp + mid_digit + temp[::-1]
                    
                else:
                    final = curr + curr[::-1]
                    
                if final[0] != '0' and int(final) % k == 0 and str(sorted(final)) not in seen:
                    seen.add(str(sorted(final)))
                    res[0] += count_unique_rearrangements_exclude_leading_zero(final)
                
                return 
            
            for c in '0123456789':
                dfs(i+1, n, curr + c)
            
        dfs(0, n, '')
        return res[0]
        
       
        
                    
                
                
            
        
25153