<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:
        pow10 = [1] * 11
        for i in range(1, 11):
            pow10[i] = pow10[i - 1] * i

        valid = []
        half = (n + 1) // 2
        low = 10 ** (half - 1)
        high = 10 ** half - 1
        
        if n == 1:
            low = 0
        
        for first_half in range(low, high + 1):
            half_str = str(first_half)
            if n % 2 == 0:
                full = half_str + half_str[::-1]
            else:
                rest = half_str[:-1]
                full = half_str + rest[::-1]

            if len(full) == n:
                if int(full) % k == 0:
                    valid.append(int(full))
        
        unique = set()
        res = 0
        for v in valid:
            vs = str(v).zfill(n)
            digit_freq = Counter(map(int, vs))
            unique.add(tuple(digit_freq.get(i, 0) for i in range(10)))

        for uniq_freq in unique:
            # print(uniq_freq)
            perm_count = 0
            for i in range(1, 10):
                if uniq_freq[i]:
                    modified = list(uniq_freq)
                    modified[i] -= 1
                    flag = True
                    for freq in modified:
                        if freq < 0:
                            flag = False
                            break
                    if not flag:
                        continue

                    vp = pow10[n - 1]
                    for i in range(10):
                        vp //= pow10[modified[i]]

                    perm_count += vp
            res += perm_count
        return res
        
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