CF1336E2 Chiori and Doll Picking (hard version)

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 $ .