<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>Return the<strong> largest</strong> integer having <code>n</code> digits (as a string) that is <strong>k-palindromic</strong>.</p>

<p><strong>Note</strong> that the integer must <strong>not</strong> have leading zeros.</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">&quot;595&quot;</span></p>

<p><strong>Explanation:</strong></p>

<p>595 is the largest k-palindromic integer with 3 digits.</p>
</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">&quot;8&quot;</span></p>

<p><strong>Explanation:</strong></p>

<p>4 and 8 are the only k-palindromic integers with 1 digit.</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">&quot;89898&quot;</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= 9</code></li>
</ul>

class Solution:
    def largestPalindrome(self, n: int, k: int) -> str:
        def nr(i, d, r): return (d * (pow(10, i, k) + pow(10, n - 1 - i, k)) + r) % k
        @cache
        def digit(i, r):
            if 2 * i + 1 == n:
                for d in range(9, 0, -1):
                    if (pow(10, i, k) * d + r) % k == 0:
                        return d
                return None
            if 2 * i >= n:
                return inf if r == 0 else None
            for d in range(9, 0, -1):
                if digit(i + 1, nr(i, d, r)):
                    return d
        res = [None] * n
        r = 0
        for i in range(n // 2):
            d = digit(i, r)
            res[i] = res[n - 1 - i] = str(d)
            r = nr(i, d, r)
        if n % 2 == 1:
            res[n // 2] = str(digit(n // 2, r))
        return ''.join(res)
        

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