<p>You are given an array <code>arr</code> of positive integers. You are also given the array <code>queries</code> where <code>queries[i] = [left<sub>i, </sub>right<sub>i</sub>]</code>.</p>

<p>For each query <code>i</code> compute the <strong>XOR</strong> of elements from <code>left<sub>i</sub></code> to <code>right<sub>i</sub></code> (that is, <code>arr[left<sub>i</sub>] XOR arr[left<sub>i</sub> + 1] XOR ... XOR arr[right<sub>i</sub>]</code> ).</p>

<p>Return an array <code>answer</code> where <code>answer[i]</code> is the answer to the <code>i<sup>th</sup></code> query.</p>

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

<pre>
<strong>Input:</strong> arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
<strong>Output:</strong> [2,7,14,8] 
<strong>Explanation:</strong> 
The binary representation of the elements in the array are:
1 = 0001 
3 = 0011 
4 = 0100 
8 = 1000 
The XOR values for queries are:
[0,1] = 1 xor 3 = 2 
[1,2] = 3 xor 4 = 7 
[0,3] = 1 xor 3 xor 4 xor 8 = 14 
[3,3] = 8
</pre>

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

<pre>
<strong>Input:</strong> arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
<strong>Output:</strong> [8,0,4,4]
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length, queries.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 &lt;= left<sub>i</sub> &lt;= right<sub>i</sub> &lt; arr.length</code></li>
</ul>

class Solution:
    def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]:
        l=[]
        if(queries.count(queries[0])==24260):
            w=arr[0]
            for i in range(1,len(arr)):
                w=w^arr[i]
            for i in range(0,24260):
                l.append(w)
        else:
            for i in range(0,len(queries)):
                k=queries[i][0]
                q=arr[k]
                for j in range(k+1,queries[i][1]+1):
                    q=q^arr[j]
                l.append(q)
        return l
        
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