The **XOR sum** of a list is the bitwise `XOR`

of all its elements. If the list only contains one element, then its **XOR sum** will be equal to this element.

- For example, the
**XOR sum**of`[1,2,3,4]`

is equal to`1 XOR 2 XOR 3 XOR 4 = 4`

, and the**XOR sum**of`[3]`

is equal to`3`

.

You are given two **0-indexed** arrays `arr1`

and `arr2`

that consist only of non-negative integers.

Consider the list containing the result of `arr1[i] AND arr2[j]`

(bitwise `AND`

) for every `(i, j)`

pair where `0 <= i < arr1.length`

and `0 <= j < arr2.length`

.

Return *the XOR sum of the aforementioned list*.

**Example 1:**

Input:arr1 = [1,2,3], arr2 = [6,5]Output:0Explanation:The list = [1 AND 6, 1 AND 5, 2 AND 6, 2 AND 5, 3 AND 6, 3 AND 5] = [0,1,2,0,2,1]. The XOR sum = 0 XOR 1 XOR 2 XOR 0 XOR 2 XOR 1 = 0.

**Example 2:**

Input:arr1 = [12], arr2 = [4]Output:4Explanation:The list = [12 AND 4] = [4]. The XOR sum = 4.

**Constraints:**

`1 <= arr1.length, arr2.length <= 10`

^{5}`0 <= arr1[i], arr2[j] <= 10`

^{9}

class Solution {
public:
int getXORSum(vector<int>& arr1, vector<int>& arr2) {
}
};