<p>Given an array of integers <code>arr</code>.</p>

<p>We want to select three indices <code>i</code>, <code>j</code> and <code>k</code> where <code>(0 &lt;= i &lt; j &lt;= k &lt; arr.length)</code>.</p>

<p>Let&#39;s define <code>a</code> and <code>b</code> as follows:</p>

<ul>
	<li><code>a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]</code></li>
	<li><code>b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]</code></li>
</ul>

<p>Note that <strong>^</strong> denotes the <strong>bitwise-xor</strong> operation.</p>

<p>Return <em>the number of triplets</em> (<code>i</code>, <code>j</code> and <code>k</code>) Where <code>a == b</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [2,3,1,6,7]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,1,1,1,1]
<strong>Output:</strong> 10
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 300</code></li>
	<li><code>1 &lt;= arr[i] &lt;= 10<sup>8</sup></code></li>
</ul>

class Solution:
    def countTriplets(self, arr: List[int]) -> int:
        '''
        Prefix Sum
        d: key = xor_num, val = count
        Formula: xor_num - 0 in d
        '''
        n, prefix = len(arr), {-1: 0}
        val = 0
        res = 0
        for i, num in enumerate(arr):
            val = val ^ num
            prefix[i] = val
        for k in range(len(arr)):
            for i in range(-1, k):
                # print(f"checking prefix {i} and {k}")
                if prefix[k] ^ prefix[i] == 0: res += (k-i-1)
        return res
16300