python3 - <<'EOF'
from functools import lru_cache
V=[[10,5,4],[10,8,6],[10,8],[10,3]]
A=[]
for i in range(4):
    for x in V[i]:
        A.append((i,x))
n=len(A)
full=(1<<n)-1
@lru_cache(None)
def f(a,b,t):
    used=a|b
    if used==full:
        s=0
        for i in range(4):
            p=0
            q=0
            for j in range(n):
                if A[j][0]==i:
                    if (a>>j)&1: p=max(p,A[j][1])
                    if (b>>j)&1: q=max(q,A[j][1])
            s=s p-q
        return s
    if t==0:
        best=-999
        for j in range(n):
            if not ((used>>j)&1):
                z=f(a|(1<<j),b,1)
                if z>best: best=z
        return best
    else:
        best=999
        for j in range(n):
            if not ((used>>j)&1):
                z=f(a,b|(1<<j),0)
                if z<best: best=z
        return best
print(f(0,0,0))
for j in range(n):
    print(j,A[j],f(1<<j,0,1))
EOF