<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:
        n1s = [1]
        for i in range(n) :
            n1s.append(n1s[-1]*10%k)
        self.to_ret = [''] * n

        @functools.lru_cache(None)
        def solve(pt=1, hist_d=0) :
            if pt * 2 > n + 1 :
                return hist_d==0
            
            for j in range(9, -1, -1) :
                if n % 2 == 1 and pt == n // 2 + 1 :
                    now_d = (hist_d + n1s[n//2] *  j) % k
                else :
                    now_d = (hist_d + (n1s[pt-1]+n1s[n-pt]) *  j) % k
                mark = solve(pt+1, now_d)
                if mark :
                    if pt == n // 2 + 1 :
                        self.to_ret[pt-1] = str(j)
                    else :
                        self.to_ret[pt-1] = self.to_ret[n-pt] = str(j)
                    return True
            return False

        solve()
        solve.cache_clear()
        return ''.join(self.to_ret)
            
        
        
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