import sys
input = sys.stdin.readline

P=int(input())
A=list(map(int,input().split()))
B=list(map(int,input().split()))

if P==2:
    print(A[0]*B[0]%P)
    exit()

# FFT
# mod=998244353　における、NTTによる高速フーリエ変換、畳み込み
# 他の提出を参考にしており、あまり理解できていません……
mod=998244353
 
Weight=[1, 998244352, 911660635, 372528824, 929031873, 452798380, 922799308, 781712469, 476477967, 166035806, 258648936, 584193783, 63912897, 350007156, 666702199, 968855178, 629671588, 24514907, 996173970, 363395222, 565042129, 733596141, 267099868, 15311432, 0]
Weight_inv=[1, 998244352, 86583718, 509520358, 337190230, 87557064, 609441965, 135236158, 304459705, 685443576, 381598368, 335559352, 129292727, 358024708, 814576206, 708402881, 283043518, 3707709, 121392023, 704923114, 950391366, 428961804, 382752275, 469870224, 0]
 
def fft(A,n,h,inverse=0):
 
    if inverse==0:
    
        for i in range(h):
            m=1<<(h-i-1)
            for j in range(1<<i):
                w=1     
                ij=j*m*2
                
                wk=Weight[h-i]
                    
                for k in range(m):
                    A[ij+k],A[ij+k+m]=(A[ij+k]+A[ij+k+m])%mod,(A[ij+k]-A[ij+k+m])*w%mod
                    w=w*wk%mod
    else:
        for i in range(h):
            m=1<<i
            for j in range(1<<(h-i-1)):
                w=1     
                ij=j*m*2
                
                wk=Weight_inv[i+1]
                    
                for k in range(m):
                    A[ij+k],A[ij+k+m]=(A[ij+k]+A[ij+k+m]*w)%mod,(A[ij+k]-A[ij+k+m]*w)%mod
                    w=w*wk%mod
        
 
    if inverse==1:
        INV_n=pow(n,mod-2,mod)
        for i in range(n):
            A[i]=A[i]*INV_n%mod
 
    return A
 
def convolution(A,B):
    
    FFTLEN=len(A)+len(B)-1
    h=FFTLEN.bit_length()
    LEN=2**h
 
    A+=[0]*(LEN-len(A)) # A,Bのサイズを2ベキに揃える
    B+=[0]*(LEN-len(B))
 
    A_FFT=fft(A,LEN,h)
    B_FFT=fft(B,LEN,h)
 
    for i in range(len(A)):
        A[i]=A[i]*B[i]%mod
 
    A=fft(A,LEN,h,1)
 
    return A[:FFTLEN]

# Pの原始根rootを求める。

# P-1の約数を求める。
x=(P-1)
L=int((P-1)**(1/2))

FACT=dict()

for i in range(2,L+2):
    while x%i==0:
        FACT[i]=FACT.get(i,0)+1
        x=x//i

if x!=1:
    FACT[x]=FACT.get(x,0)+1

# 2から順にチェック
for i in range(2,P):
    for f in FACT:
        if pow(2,(P-1)//f,P)==1:
            break
    else:
        root=i
        break


LIST=[1]
for i in range(P):
    LIST.append(LIST[-1]*root%P)

A2=[]
B2=[]

for i in range(P-1):
    A2.append(A[LIST[i]-1])
    B2.append(B[LIST[i]-1])

#print(A)
#print(A2)

X=convolution(A2,B2)

ANS=[0]*(P-1)

for i in range(len(X)):
    k=i%(P-1)
    ANS[k]=ANS[k]+X[i]

#print(*ANS)

ANS2=[0]*(P-1)

for i in range(P-1):
    ANS2[LIST[i]-1]=ANS[i]

print(*ANS2)