Problem:
Score: 500 points

Problem Statement

Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.

In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:

A team should assign one member to each food, and should not assign the same member to multiple foods.

It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.

The score of a team is the longest time it takes for an individual member to finish the food.

Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.

What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

0 \leq K \leq 10^{18}

1 \leq A_i \leq 10^6\ (1 \leq i \leq N)

1 \leq F_i \leq 10^6\ (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N

Output

Print the minimum possible score of the team.

Sample Input 1

3 5
4 2 1
2 3 1

Sample Output 1

2

They can achieve the score of 2, as follows:

Member 1 does 4 sets of training and eats Food 2 in (4-4) \times 3 = 0 seconds.

Member 2 does 1 set of training and eats Food 3 in (2-1) \times 1 = 1 second.

Member 3 does 0 sets of training and eats Food 1 in (1-0) \times 2 = 2 seconds.

They cannot achieve a score of less than 2, so the answer is 2.

Sample Input 2

3 8
4 2 1
2 3 1

Sample Output 2

0

They can choose not to do exactly K sets of training.

Sample Input 3

11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2

Sample Output 3

12

Solution in Go (1.6) (judge verdict: Time Limit Exceeded)

package main

import (
	"bufio"
	"container/heap"
	"fmt"
	"os"
	"sort"
	"strconv"
)

var _sw = bufio.NewScanner(os.Stdin)

func main() {
	N := NextInt()
	K := NextInt()
	A := NextInts(N)
	F := NextInts(N)

	ReverseSortInts(A)
	SortInts(F)

	pq := make(PriorityQueue, N)
	for i := 0; i < N; i++ {
		pq[i] = &Item{
			value:    i,
			priority: A[i] * F[i],
			index:    i,
		}
	}
	heap.Init(&pq)

	for i := 0; i < K; i++ {
		item := pq[0]
		if item.priority > 0 {
			pq.update(item, item.value, item.priority-F[item.value])
		}
	}

	fmt.Println(heap.Pop(&pq).(*Item).priority)
}

func init() {
	_sw.Split(bufio.ScanWords)
}

func NextInt() int {
	if _sw.Scan() == false {
		panic(_sw.Err())
	}
	i, e := strconv.Atoi(_sw.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func NextInts(n int) []int {
	a := make([]int, n)
	for i := range a {
		a[i] = NextInt()
	}
	return a
}

func NextString() string {
	if _sw.Scan() == false {
		panic(_sw.Err())
	}
	return _sw.Text()
}

func Max(x, y int) int {
	if x > y {
		return x
	}
	return y
}

func Min(x, y int) int {
	if x > y {
		return y
	}
	return x
}

func Bounded(lower, n, upper int) int {
	return Min(upper, Max(lower, n))
}

func SortInts(xs []int) {
	sort.Sort(sort.IntSlice(xs))
}

func ReverseSortInts(xs []int) {
	sort.Sort(sort.Reverse(sort.IntSlice(xs)))
}

// An IntHeap is a min-heap of ints.
type IntHeap []int

func (h IntHeap) Len() int           { return len(h) }
func (h IntHeap) Less(i, j int) bool { return h[i] < h[j] }
func (h IntHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *IntHeap) Push(x interface{}) {
	// Push and Pop use pointer receivers because they modify the slice's length,
	// not just its contents.
	*h = append(*h, x.(int))
}

func (h *IntHeap) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}

// An Item is something we manage in a priority queue.
type Item struct {
	value    int // The value of the item; arbitrary.
	priority int // The priority of the item in the queue.
	// The index is needed by update and is maintained by the heap.Interface methods.
	index int // The index of the item in the heap.
}

// A PriorityQueue implements heap.Interface and holds Items.
type PriorityQueue []*Item

func (pq PriorityQueue) Len() int { return len(pq) }

func (pq PriorityQueue) Less(i, j int) bool {
	// We want Pop to give us the highest, not lowest, priority so we use greater than here.
	return pq[i].priority > pq[j].priority
}

func (pq PriorityQueue) Swap(i, j int) {
	pq[i], pq[j] = pq[j], pq[i]
	pq[i].index = i
	pq[j].index = j
}

func (pq *PriorityQueue) Push(x interface{}) {
	n := len(*pq)
	item := x.(*Item)
	item.index = n
	*pq = append(*pq, item)
}

func (pq *PriorityQueue) Pop() interface{} {
	old := *pq
	n := len(old)
	item := old[n-1]
	old[n-1] = nil  // avoid memory leak
	item.index = -1 // for safety
	*pq = old[0 : n-1]
	return item
}

// update modifies the priority and value of an Item in the queue.
func (pq *PriorityQueue) update(item *Item, value int, priority int) {
	item.value = value
	item.priority = priority
	heap.Fix(pq, item.index)
}