#!/usr/local/bin/python3
def rk(M,n):
 M=[r[:] for r in M];r=0
 for c in range(n):
  p=next((i for i in range(r,n) if M[i][c]),-1)
  if p<0:continue
  M[r],M[p]=M[p],M[r]
  for i in range(r+1,n):
   if M[i][c]:
    for j in range(c,n):M[i][j]^=M[r][j]
  r+=1
 return r
def dt(M,n):return rk(M,n)==n
def iv(M,n):
 A=[M[i][:]+[int(i==j) for j in range(n)] for i in range(n)];r=0
 for c in range(n):
  p=next((i for i in range(r,n) if A[i][c]),-1)
  if p<0:return None
  A[r],A[p]=A[p],A[r]
  for i in range(n):
   if i!=r and A[i][c]:
    for j in range(c,2*n):A[i][j]^=A[r][j]
  r+=1
 return [q[n:] for q in A]
t=4;n=6
E=[(i,j) for i in range(t) for j in range(i+1,t)]
Ws=[]
for m in range(1<<6):
 W=[[0]*t for _ in range(t)]
 for k,(i,j) in enumerate(E):
  if m>>k&1:W[i][j]=W[j][i]=1
 if dt(W,t):Ws.append(W)
for bits in range(1<<t):
 if bits in (0,15):continue
 L=[[(bits>>j)&1,1-((bits>>j)&1)] for j in range(t)]
 # basis of kernel L^T dim2
 U=[]
 for free in range(t):
  v=[0]*t;v[free]=1
  # need L^T v=0 i.e. for each col a sum v_i L_i[a]=0; set pivots solve simply by try all?
  pass
 # brute all v nonzero kernel
 ker=[]
 for v in range(1,1<<t):
  ok=True
  for a in range(2):
   if sum(((v>>i)&1)*L[i][a] for i in range(t))% 2:ok=False
  if ok:ker.append([(v>>i)&1 for i in range(t)])
 # ker should have 3 nonzero vectors
 q1=0;q2=0
 for W in Ws:
  F=iv(W,t)
  z=True
  for u in ker:
   for v in ker:
    if sum(u[i]*F[i][j]*v[j] for i in range(t) for j in range(t))% 2:z=False
  if z:q1+=1
  # actual A det
  K=[[(F[i][j]^(i==j)^(sum(L[i][a]*L[j][a] for a in range(2))% 2)) for j in range(t)] for i in range(t)]
  A=[[0]*n for _ in range(n)]
  for i in range(2):
   for j in range(t):A[i][2+j]=A[2+j][i]=L[j][i]
  for i in range(t):
   for j in range(t):A[2+i][2+j]=K[i][j]
  if dt(A,n):q2+=1
 print(bits,q1,q2)