<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:
        D0 = {dd: (sorted(set((dd * xx) % 10 for xx in range(1, 11)), reverse=True)) for dd in range(1, 10)}
        if k == 1:
            return "9" * n

        @functools.lru_cache(None)
        def dp(i, modk):
            if i >= ((n + 1) // 2):
                if modk == 0:
                    return 99
                else:
                    return -1
            if i == 0:
                D = D0[k]
            else:
                D = range(9, -1, -1)
            for d in D:
                if (n - i - 1) != i:
                    ppp = pow(10, i, k) + pow(10, n - i - 1, k)
                else:
                    ppp = pow(10, i, k)
                modk2 = (modk + d * ppp) % k
                if dp(i + 1, modk2) >= 0:
                    return d
            return -1

        modk = 0
        sol = []
        for i in range((n + 1) // 2):
            d = dp(i, modk)
            ppp = pow(10, i, k) + pow(10, n - i - 1, k)
            modk = (modk + d * ppp) % k
            sol.append(str(d))
        for i in range(n // 2):
            sol.append(sol[n // 2 - i - 1])
        dp.cache_clear()
        return ''.join(sol)

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