if k is odd, we pick the largest k//2 nums and the smallest k//2 nums, depending on the sign of B[i], either largest -smallest or smallest-largest

if k is even, we either pick k//2-1 largest and k//2 smallest, or k//2 largest and k//2-1 smallest

```
A=[1,4,-3,4,-6,-4,6,7,8]
B=[1,2,-4,5,-6,-7,2,1,2]
k=3
low,high=[],[] # store the largest k//2 A[i] and the smallest k//2 A[i]
th=tl=0 # sum of the largest k//2 A[i] and sum of the smallest k//2 A[i]
res=float('-inf')
for b,a in sorted(zip(B,A),reverse=True): # sort pairs by B from high to low
if k%2:
res=max([res, (th-tl+a)*b,(tl-th+a)*b]) # if b is negative, smallest minus the largest; else, largest-smallest
elif high and low:
i,j=th-high[0]+a-tl,tl+low[0]+a-th
res=max([res,i*b,-i*b,j*b,-j*b ])
heapq.heappush(high,a)
heapq.heappush(low,-a)
th+=a
tl-=-a
if len(high)>k//2:
th-=heapq.heappop(high)
if len(low)>k//2:
tl+=heapq.heappop(low)
return res
```