Chiori and Doll Picking (easy version)

This is the easy version of the problem. The only difference between easy and hard versions is the constraint of $$$m$$$. You can make hacks only if both versions are solved.

Chiori loves dolls and now she is going to decorate her bedroom!

As a doll collector, Chiori has got $$$n$$$ dolls. The $$$i$$$-th doll has a non-negative integer value $$$a_i$$$ ($$$a_i < 2^m$$$, $$$m$$$ is given). Chiori wants to pick some (maybe zero) dolls for the decoration, so there are $$$2^n$$$ different picking ways.

Let $$$x$$$ be the bitwise-xor-sum of values of dolls Chiori picks (in case Chiori picks no dolls $$$x = 0$$$). The value of this picking way is equal to the number of $$$1$$$-bits in the binary representation of $$$x$$$. More formally, it is also equal to the number of indices $$$0 \leq i < m$$$, such that $$$\left\lfloor \frac{x}{2^i} \right\rfloor$$$ is odd.

Tell her the number of picking ways with value $$$i$$$ for each integer $$$i$$$ from $$$0$$$ to $$$m$$$. Due to the answers can be very huge, print them by modulo $$$998\,244\,353$$$.