Let us consider two cases:

• Every number appears an even number of times in the array.

  In this case, no matter which number $$$x$$$ we choose, the array will contain an even number of its submasks. Therefore, the answer is -1.

• There exists a number that appears an odd number of times in the array.

  Let $$$x$$$ be the smallest such number. Note that if $$$y$$$ is a submask of $$$x$$$, then $$$y \le x$$$. All numbers smaller than $$$x$$$ appear an even number of times, and the number $$$x$$$ itself appears an odd number of times. Thus, $$$x$$$ is a valid solution.