This is the hard 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!

$\text{[math]}$

As a doll collector, Chiori has got $$$n$$$ dolls. The $$$i$$$-th doll has a non-negative integer value $$$a_i$$$ ($$$a_i \lt 2^m$$$, $$$m$$$ 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 $$$0 \leq i \lt m$$$, 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 $$$998\,244\,353$$$.

Input

The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$0 \le m \le 53$$$) — the number of dolls and the maximum value of the picking way.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \lt 2^m$$$) — the values of dolls.

Output

Print $$$m+1$$$ integers $$$p_0, p_1, \ldots, p_m$$$ — $$$p_i$$$ is equal to the number of picking ways with value $$$i$$$ by modulo $$$998\,244\,353$$$.

Chiori and Doll Picking (hard version)

Examples