<p>Given an integer array nums, return <em>the number of <strong>AND triples</strong></em>.</p>

<p>An <strong>AND triple</strong> is a triple of indices <code>(i, j, k)</code> such that:</p>

<ul>
	<li><code>0 &lt;= i &lt; nums.length</code></li>
	<li><code>0 &lt;= j &lt; nums.length</code></li>
	<li><code>0 &lt;= k &lt; nums.length</code></li>
	<li><code>nums[i] &amp; nums[j] &amp; nums[k] == 0</code>, where <code>&amp;</code> represents the bitwise-AND operator.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,1,3]
<strong>Output:</strong> 12
<strong>Explanation:</strong> We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 &amp; 2 &amp; 1
(i=0, j=1, k=0) : 2 &amp; 1 &amp; 2
(i=0, j=1, k=1) : 2 &amp; 1 &amp; 1
(i=0, j=1, k=2) : 2 &amp; 1 &amp; 3
(i=0, j=2, k=1) : 2 &amp; 3 &amp; 1
(i=1, j=0, k=0) : 1 &amp; 2 &amp; 2
(i=1, j=0, k=1) : 1 &amp; 2 &amp; 1
(i=1, j=0, k=2) : 1 &amp; 2 &amp; 3
(i=1, j=1, k=0) : 1 &amp; 1 &amp; 2
(i=1, j=2, k=0) : 1 &amp; 3 &amp; 2
(i=2, j=0, k=1) : 3 &amp; 2 &amp; 1
(i=2, j=1, k=0) : 3 &amp; 1 &amp; 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,0,0]
<strong>Output:</strong> 27
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt; 2<sup>16</sup></code></li>
</ul>

def fwht(x):
    n, m, A = len(x), 1, x.copy()
    while 2 * m <= n:
        for i in range(0, n, 2 * m):
            for j in range(i, i + m):
                x, y = A[j], A[j + m]
                A[j], A[j + m] = y, x + y
        m *= 2
    return A

def ifwht(x):
    n, m, A = len(x), 1, x.copy()
    while 2 * m <= n:
        for i in range(0, n, 2 * m):
            for j in range(i, i + m):
                x, y = A[j], A[j + m]
                A[j], A[j + m] = -x + y, x
                break
        m *= 2
    return A

class Solution:
    def countTriplets(self, A: 'List[int]') -> 'int':
        n, M, cnt = 1, max(A), collections.Counter(A)
        while n <= M: n *= 2
        v = [cnt[i] for i in range(n)]
        v = ifwht([x**3 for x in fwht(v)])
        return v[0]
18900