<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:
        n = len(arr)
        preXorSum = [0] * (n + 1)
        for i in range(n):
            preXorSum[i + 1] = preXorSum[i] ^ arr[i]
        ans = 0
        for i in range(n):
            for j in range(i + 1, n):
                for k in range(j, n):
                    if preXorSum[j] ^ preXorSum[i] == preXorSum[k + 1] ^ preXorSum[j]:
                        ans += 1
        return ans
1563