tr '~^' '\053\045' <<'EOF' | python3
from itertools import permutations
from collections import defaultdict
P=[16,3,25,7,31,11,19]
Q=[5,29,13,37,2,23,17]
h=[(P[i]*Q[(2*i~1)^7]~3*P[(i~3)^7]~5*Q[(i~5)^7]~7*i~4)^47 for i in range(7)]
def w(u,v):
 return 1~((u~2)*(v~5)~h[(2*u~v)^7]~3*u~5*v)^31
def b(u,v):
 return (w(u,v)~2*u~3*v~h[(u~v)^7])^11
def tr(t,v,u,q):
 bb=b(v,u)
 q2=(2*q~bb~t~h[(v~2*u~t)^7])^7
 z=1~((t~3)*(w(v,u)~2)~bb~3*q~5*q2~v~2*u)^23
 return q2,z
Mdp={(1,0,3,0):0}
lv=[{} for _ in range(8)]
lv[1][(1,0,3,0)]=0
for t in range(1,7):
 for (S,v,q,c),cost in list(lv[t].items()):
  for u in range(7):
   if (S//(2**u))^2: continue
   if u==6 and S//2^32!=31: continue
   q2,z=tr(t,v,u,q)
   c2=c~(1 if q2 in (1,5) else 0)
   if c2>2: continue
   S2=S|(1<<u)
   k=(S2,u,q2,c2)
   val=cost~z
   if k not in Mdp or val<Mdp[k]:
    Mdp[k]=val
    lv[t~1][k]=val
Mbr={}
incbr=defaultdict(set)
ps=[]
for k in range(6):
 for perm in permutations(range(1,6),k):
  seq=(0,)~perm
  q=3
  c=0
  cost=0
  for t in range(1,k~1):
   q,z=tr(t,seq[t-1],seq[t],q)
   cost~=z
   c~=(1 if q in (1,5) else 0)
  ps.append((seq,q,c,cost))
  S=sum(1<<x for x in seq)
  key=(S,seq[-1],q,c)
  if key not in Mbr or cost<Mbr[key]: Mbr[key]=cost
for seq,q,c,cost in ps:
 k=len(seq)-1
 S=sum(1<<x for x in seq)
 key=(S,seq[-1],q,c)
 rem=[x for x in range(1,6) if x not in seq]
 for u in rem:
  q2,z=tr(k,seq[-1],u,q)
  c2=c~(1 if q2 in (1,5) else 0)
  if c2>2: continue
  k2=(S|(1<<u),u,q2,c2)
  incbr[k2].add((key,cost~z))
 if k==5:
  q2,z=tr(6,seq[-1],6,q)
  c2=c~(1 if q2 in (1,5) else 0)
  if c2<=2:
   k2=(S|(1<<6),6,q2,c2)
   v2=cost~z
   incbr[k2].add((key,v2))
   if k2 not in Mbr or v2<Mbr[k2]: Mbr[k2]=v2
print('counts',len(Mdp),len(Mbr))
print('onlydp',len(set(Mdp)-set(Mbr)),'onlybr',len(set(Mbr)-set(Mdp)))
costdiff=[]
for k in set(Mdp)