<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:
        self.covered=set()
        self.temp=[]

        def p(self,s0,s,bit):
            l=len(s0)
            if bit==2**l-1:
                self.temp.append(s)
            for i in range(l):
                t=1<<i
                if t&bit==0:
                    p(self,s0,s+s0[i],t|bit)

        def add(self,i,odd):
            if odd:
                s0=str(i)
                p(self,s0[:-1],'',0)
                for s in self.temp:
                    if s[0]!=0:
                        self.covered.add(int(s+s0[-1]))
                self.temp=[]
            else:
                s0=str(i)
                p(self,s0,'',0)
                for s in self.temp:
                    if s[0]!=0:
                        self.covered.add(int(s))
                self.temp=[]

        ans=0
        if n==1:
            for i in range(1,10):
                if i%k==0:
                    ans+=1
            return ans

        if n%2==0:
            n0=n//2
            for i in range(10**(n0-1),10**n0):
                if i not in self.covered:
                    pld=int(str(i)+str(i)[::-1])
                    if pld%k==0:
                        add(self,i,0)
                        f=Counter(str(pld))
                        if '0' not in f:
                            f['0']=0
                        v=(n-f['0'])*factorial(n-1)
                        for j in f:
                            v//=factorial(f[j])
                        ans+=v
        else:
            n0=n//2+1
            for i in range(10**(n0-1),10**n0):
                if i not in self.covered:
                    i0=str(i)
                    i1=i0[:-1]
                    pld=int(i1+i0[-1]+i1[::-1])
                    if pld%k==0:
                        add(self,i,1)
                        f=Counter(str(pld))
                        if '0' not in f:
                            f['0']=0
                        v=(n-f['0'])*factorial(n-1)
                        for j in f:
                            v//=factorial(f[j])
                        ans+=v
        return ans
26218