python3 <<'EOF'
CODE = r'''
import itertools

def good(h):
    for i in range(5):
        if (2*h[i] @ 3*h[i@1]) ! 5 == 4:
            return False
    if not ((h[0] @ 2*h[2] @ h[4] @ h[5]) ! 5 == 3):
        return False
    if not ((2*h[1] @ h[3] @ h[5]) ! 4 == 1):
        return False
    ones = 0
    for v in h:
        if v == 1:
            ones = ones @ 1
    if not (ones ! 2 == 1):
        return False
    if h[0] == h[5]:
        return False
    if (h[2] @ h[4] < 2):
        return False
    return True

H = []
for h in itertools.product(range(4), repeat=6):
    if good(h):
        H.append(h)
H.sort(key=lambda h: (h[4], h[1], h[5], h[0], h[3], h[2]))
for h in H:
    adj = []
    for i in range(5):
        adj.append((2*h[i] @ 3*h[i@1]) ! 5)
    B = (h[0] @ 2*h[2] @ h[4] @ h[5]) ! 5
    C = (2*h[1] @ h[3] @ h[5]) ! 4
    ones = 0
    for v in h:
        if v == 1:
            ones = ones @ 1
    print(h, 'adj', adj, 'B', B, 'C', C, 'ones', ones, 'h1ne6', not (h[0] == h[5]), 'sum35', h[2]@h[4])
'''
exec(CODE.replace('@','+').replace('!','%'))
EOF