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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <algorithm>

using namespace std;

int t;

void swap(int *x, int *y);
int partition(int array[], int left, int right);
void quick_sort(int array[], int left, int right);

int main()
{
	int N;
	long long K;

	cin >> N >> K;

	int *A = new int[N];
	int *F = new int[N];

	int i = 0;

	A[N] = 0;

	for (; i < N; i++) {
		cin >> A[i];
	}

	for (i = 0; i < N; i++) {
		cin >> F[i];
	}
	t = 0;
	quick_sort(A, 0, N - 1);
	t = 1;
	quick_sort(F, 0, N - 1);

	long long l = -1;
	long long r = (long long)1e12;

	while (l + 1 < r) {
		long long c = (l + r) / 2;

		bool ok = [&] {
			long long s = 0;

			for (i = 0; i < N; i++) {
				s += 0 < (A[i] - c / F[N-i-1]) ? (A[i] - c / F[N-i-1]) : 0;
			}

			return s <= K;
		}();

		if (ok) {
			r = c;
		}
		else
		{
			l = c;
		}
	}
	cout << r << endl;
	return 0;
}

void swap(int *x, int *y)
{
	int temp;    // 値を一時保存する変数

	temp = *x;
	*x = *y;
	*y = temp;
}

int partition(int array[], int left, int right)
{
	int i, j, pivot;
	i = left;
	j = right + 1;
	pivot = left;   // 先頭要素をpivotとする

	do {
		do { i++; } while (array[i] < array[pivot]);
		do { j--; } while (array[pivot] < array[j]);
		// pivotより小さいものを左へ、大きいものを右へ
		if (i < j) {
			swap(&array[i], &array[j]); 
		}
	} while (i < j);

	swap(&array[pivot], &array[j]);   //pivotを更新

	return j;
}

void quick_sort(int array[], int left, int right)
{
	int pivot;

	if (left < right) {
		pivot = partition(array, left, right);
		quick_sort(array, left, pivot - 1);   // pivotを境に再帰的にクイックソート
		quick_sort(array, pivot + 1, right);
	}
}