<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>

class Solution:
    def countTriplets(self, A: 'List[int]') -> 'int':
	
        tmp = []
        maxlen = 0
        for a in A:
            tmp.append(bin(a)[2:])
            maxlen = max(maxlen, len(tmp[-1]))
        pool = []
        for s in tmp:
            extra = maxlen - len(s)
            pool.append('0'*extra + s)
        
        row, col = len(pool), len(pool[0])
        one = collections.defaultdict(set)
        for j in range(col):
            for i in range(row):
                if pool[i][j] == '1':
                    one[j].add(i)
        
        Venn = collections.defaultdict(list)
        cnt = 0
        for j in range(col):
            if len(one[j]) != 0:
                cnt += (len(one[j]))**3
                for i in range(j, 0, -1):
                    for prv in Venn[i]:
                        intersec = prv & one[j]
                        if len(intersec) != 0:
                            cnt += ((-1)**i)*(len(intersec))**3
                            Venn[i+1].append(intersec)
                Venn[1].append(one[j])
        
        return row**3 - cnt
69100