QUESTION:
You are given an array A of length N. You can perform the following operation on it:

Pick 2 distinct indices i and j, remove A[i] and A[j] from A, then push (A[i] xor A[j]) at the back of A.
Let's define F(A) as the minimum sum of elements of A after performing the above operation on A as many times as you want.
You have to remove exactly 1 element from A such that the value of F(A) is minimized.

Note:

• You pass the array A after removing the element in F(A).
• XOR of empty array is 0.

Determine the minimum value of F(A) you can achieve by removing exactly 1 element from A.

Example:
Given,
N = 3
A = [1, 1, 2]
Approach

By removing 1st element from A, the array will become [1, 2], then the minimum sum will be 1+2=3 as there is no need to apply any operation on it.
By removing 2nd element from A, the array will become [1, 2], then the minimum sum will be 1+2=3 as there is no need to apply any operation on it.
By removing 3rd element from A, the array will become [1, 1], then apply operation combining these 2 elements, then the array will become [1^1]=[0]. So, the sum will be 0.
So, the answer will be min(3, 3, 0) = 0.

Example:
Given
N = 2
A = [1, 2]
Approach

By removing 1st element from A, the array will become [2], then the minimum sum will be 2.
By removing 2nd element from A, the array will become [1], then the minimum sum will be 1.
So, the answer will be min(2, 1)=1.

Example:
Given,

N = 3
A = [0,1,3]
Approach

By removing 1st element from A, the array will become [1, 3], then apply operation combining these 2 elements, then the array will become [1^3]=[2]. So, the sum will be 2.
By removing 2nd element from A, the array will become [0, 3], then the minimum sum will be 0+3=3 as there is no need to apply any operation on it.
By removing 3rd element from A, the array will become [0, 1], then minimum sum will be 0+1=1 as there is no need to apply any operation on it.
So, the answer will be min(2, 3, 1)=1.