<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:

        visited = set()
        
        factorial = {1:1,2:2,3:6,4:24,5:120,6:720,7:5040,8:40320,9:362880,10:3628800}
        

        if n%2==0:
            M = n//2

            if M == 1:
                left,right = 1,9
            elif M == 2:
                left,right = 10,99
            elif M == 3:
                left,right = 100,999
            elif M == 4:
                left,right = 1000,9999
            elif M == 5:
                left,right = 10000,99999
            res = 0
            for i in range(left,right+1):
                tmp_left = i*(10**M)
                s = str(i)
                tmp_right = int(str(tmp_left)[::-1])
                tmp = tmp_left+tmp_right
                if tmp % k == 0:
                    
                    t = {}
                    for j in range(len(s)):
                        if s[j] not in t.keys():
                            t[s[j]] =0
                        t[s[j]] +=2
                    
                    A = tuple(sorted(t.items()))
                    if A not in visited:
                        visited.add(A)

                        C = factorial[n]
                        for key in t.keys():
                            C = C/factorial[t[key]]
                        if '0' in t.keys():
                            C2 = factorial[n-1]
                            for key in t.keys():
                                if key != '0':
                                    C2 = C2/factorial[t[key]]
                                else:
                                    if t[key]>1:
                                        C2 = C2/factorial[t[key]-1]
                                    
                        else:
                                    C2 = 0
                        res += (C-C2)
        
        else:
            M = (n-1)//2
            if M == 0:
                res = 0
                for i in range(1,10):
                    if i %k == 0:
                        res +=1
            else:
                if M == 1:
                    left,right = 1,9
                elif M == 2:
                    left,right = 10,99
                elif M == 3:
                    left,right = 100,999
                elif M == 4:
                    left,right = 1000,9999
                res = 0            
                for p in range(0,10):
                    for i in range(left,right+1):
                        
                        tmp_left = i*(10**(M+1))
                        s = str(i)
                        tmp_right = int(str(tmp_left)[::-1])
                        tmp_mid = p*(10**(M))
                        tmp = tmp_left+tmp_mid+tmp_right
                        if tmp %k==0:
                            
                            t = {}
                            for j in range(len(s)):
                                if s[j] not in t.keys():
                                    t[s[j]] =0
                                t[s[j]] +=2
                            if str(p) not in t.keys():
                                t[str(p)] =0
                            t[str(p)] +=1

                            A = tuple(sorted(t.items()))
                            if A not in visited:
                                visited.add(A)
                                C = factorial[n]
                                for key in t.keys():
                                    C = C/factorial[t[key]]
                                if '0' in t.keys():
                                    C2 = factorial[n-1]
                                    for key in t.keys():
                                        if key != '0':
                                            C2 = C2/factorial[t[key]]
                                        else:
                                            if t[key]>1:
                                                C2 = C2/factorial[t[key]-1]
                                else:
                                    C2 = 0
                                #print(s,tmp,t,C,C2,C-C2)
                                res += (C-C2)
            
        return int(res)

        

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