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!

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 $998\,244\,353$.

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 $a_1, a_2, \ldots, a_n$ ($0 \le a_i < 2^m$)  — the values of dolls.

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 $998\,244\,353$.