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: Wrong Answer)

#include <stdio.h>
#include <sstream>
#include <string.h>
#include <vector>
#include <map>
#include <algorithm>
#include <utility>
#include <set>
#include <cctype>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <limits>
#include <iomanip>
#include <ctype.h>
#include <unordered_map>
#include <random>
#define _USE_MATH_DEFINES
#include <iostream>
#include <math.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<int, char> pic;
#define bit(x,v) ((ll)x << v)

const ll INF = 1000000007;
const int MAX = 210000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}
int pr[100010];
void uini(int n) {
	for (size_t i = 0; i <= n; i++)
	{
		pr[i] = i;
	}
}

int parent(int x) {
	if (x == pr[x]) return x;
	return pr[x] = parent(pr[x]);
}

int same(int x, int y) {
	return parent(x) == parent(y);
}

bool unit(int x, int y) {
	int px = parent(x);
	int py = parent(y);

	if (px == py) return false;
	if (px < py) {
		pr[py] = px;
	}
	else {
		pr[px] = py;
	}
	return true;
}

ll bit[200010];
int max_n = 200000;
int pm = 0;
void add(int x) {

	while (max_n >= x)
	{
		bit[x]++;
		x += x & -x;
	}
}
void sub(int x) {
	while (max_n >= x)
	{
		bit[x]--;
		x += x & -x;
	}
}
ll merge(ll* a, int left, int mid, int right) {
	ll n1 = mid - left;
	ll n2 = right - mid;
	vector<int> L(n1 + 1);
	vector<int> R(n2 + 1);
	for (size_t i = 0; i < n1; i++)
	{
		L[i] = a[left + i];
	}
	for (size_t i = 0; i < n2; i++)
	{
		R[i] = a[mid + i];
	}

	L[n1] = INF;
	R[n2] = INF;
	ll i = 0;
	ll j = 0;
	ll r = 0;
	for (size_t k = left; k < right; k++)
	{
		if (L[i] <= R[j]) {
			a[k] = L[i];
			i++;
		}
		else {
			a[k] = R[j];
			r += n1 - i;
			j++;
		}
	}
	return r;
}
ll merge2(pair<int, char>* a, int left, int mid, int right) {
	ll n1 = mid - left;
	ll n2 = right - mid;
	vector<pair<int, char>> L(n1 + 1);
	vector<pair<int, char>> R(n2 + 1);
	for (size_t i = 0; i < n1; i++)
	{
		L[i] = a[left + i];
	}
	for (size_t i = 0; i < n2; i++)
	{
		R[i] = a[mid + i];
	}

	L[n1] = make_pair(INF, ' ');
	R[n2] = make_pair(INF, ' ');
	ll i = 0;
	ll j = 0;
	ll r = 0;
	for (size_t k = left; k < right; k++)
	{
		if (L[i].first <= R[j].first) {
			a[k] = L[i];
			i++;
		}
		else {
			a[k] = R[j];
			r += n1 - i;
			j++;
		}
	}
	return r;
}
ll mergeSort2(pair<int, char>* a, int left, int right) {
	ll res = 0;
	if (left + 1 < right) {
		int mid = (left + right) / 2;
		res = mergeSort2(a, left, mid);
		res += mergeSort2(a, mid, right);
		res += merge2(a, left, mid, right);
	}
	return res;
}
ll mergeSort(ll* a, int left, int right) {
	ll res = 0;
	if (left + 1 < right) {
		int mid = (left + right) / 2;
		res = mergeSort(a, left, mid);
		res += mergeSort(a, mid, right);
		res += merge(a, left, mid, right);
	}
	return res;
}
int partition(pair<int, char>* a, int p, int r) {
	pair<int, char> x = a[r];
	int i = p - 1;
	for (size_t j = p; j < r; j++)
	{
		if (a[j].first <= x.first) {
			i++;
			swap(a[i], a[j]);
		}
	}
	swap(a[i + 1], a[r]);
	return i + 1;
}
void quick(pair<int, char>* a, int p, int r) {
	if (p < r) {
		int q = partition(a, p, r);
		quick(a, p, q - 1);
		quick(a, q + 1, r);
	}
}

ll n;
int ci = 0;
ll P[1000010];
struct Node {
	int key;
	int priority;
	Node* parent, * left, * right;
	Node(int key, int priority);
	Node() {}
};
Node NIL;
Node::Node(int key, int priority) : key(key), priority(priority) {
	left = &NIL;
	right = &NIL;
}
Node* root = new Node();
void cenrec(Node* k) {
	if (k->key == NIL.key) return;
	cenrec(k->left);
	cout << " " << k->key;
	cenrec(k->right);
}
void fastrec(Node* k)
{
	if (k->key == NIL.key) return;
	cout << " " << k->key;
	fastrec(k->left);
	fastrec(k->right);
}
void insert(Node* v) {
	Node* y = &NIL;
	Node* x = root;
	while (x->key != NIL.key)
	{
		y = x;
		if (v->key < x->key) {
			x = x->left;
		}
		else {
			x = x->right;
		}
	}
	v->parent = y;
	if (y->key == NIL.key) {
		root = v;
	}
	else if (v->key < y->key) {
		y->left = v;
	}
	else {
		y->right = v;
	}

}

Node* find(Node* k, ll v)
{
	if (k->key == NIL.key) return &NIL;
	if (k->key == v) return k;
	if (v < k->key) return find(k->left, v);
	return find(k->right, v);
}
void delp12(Node* x) {
	if (x->key == NIL.key)  return;
	Node* l = x->left;
	Node* r = x->right;
	Node* pr = x->parent;

	if (l->key == NIL.key
		&& r->key == NIL.key) {
		if (pr->left == x) {
			pr->left = &NIL;
		}
		else pr->right = &NIL;
	}
	else if (l->key != NIL.key) {
		if (pr->left == x) {
			pr->left = l;
		}
		else pr->right = l;
		l->parent = pr;
	}
	else if (r->key != NIL.key) {
		if (pr->left == x) {
			pr->left = r;
		}
		else pr->right = r;
		r->parent = pr;
	}
}
Node* get_next(Node* k) {
	if (k->key == NIL.key) return &NIL;
	Node* res = get_next(k->left);
	if (res->key != NIL.key) return res;
	return k;
}
void del(Node* x) {

	if (x->key == NIL.key) return;
	Node* l = x->left;
	Node* r = x->right;
	Node* pr = x->parent;

	if (l->key != NIL.key && r->key != NIL.key) {
		Node* nex = get_next(r);
		x->key = nex->key;
		delp12(nex);
	}
	else {
		delp12(x);
	}
}
Node* rightRotate(Node* t) {
	Node* s = t->left;
	t->left = s->right;
	s->right = t;
	return s;
}
Node* leftRotate(Node* t) {
	Node* s = t->right;
	t->right = s->left;
	s->left = t;
	return s;
}
Node* _insert(Node* t, int key, int priority) {
	if (t->key == NIL.key) {
		return new Node(key, priority);
	}
	if (key == t->key) {
		return t;
	}

	if (key < t->key) {
		t->left = _insert(t->left, key, priority);
		if (t->priority < t->left->priority) {
			t = rightRotate(t);
		}
	}
	else {
		t->right = _insert(t->right, key, priority);
		if (t->priority < t->right->priority) {
			t = leftRotate(t);
		}
	}
	return t;
}
Node* delete1(Node* t, int key);
Node* _delete(Node* t, int key) {
	if (t->left->key == NIL.key && t->right->key == NIL.key) {
		return &NIL;
	}
	else if (t->left->key == NIL.key) {
		t = leftRotate(t);
	}
	else if (t->right->key == NIL.key) {
		t = rightRotate(t);
	}
	else
	{
		if (t->left->priority > t->right->priority) {
			t = rightRotate(t);
		}
		else
			t = leftRotate(t);
	}
	return delete1(t, key);
}
Node* delete1(Node* t, int key) {
	if (t->key == NIL.key) {
		return &NIL;
	}
	if (key < t->key) {
		t->left = delete1(t->left, key);
	}
	else if (key > t->key) {
		t->right = delete1(t->right, key);
	}
	else return _delete(t, key);
	return t;
}
int H;
int left(int i) {
	return i * 2 + 1;
}
int right(int i) {
	return i * 2 + 2;
}
struct edge {
	int from, to;
	ll val;
	edge(int from,int to,ll val) : from(from),to(to),val(val){}
};
int k;
int _rank[1010];
int temp[1010];
bool compare_sa(int i, int j) {
	if (_rank[i] != _rank[j]) return _rank[i] < _rank[j];
	else {
		int ri = i + k <= n ? _rank[i + k] : -1;
		int rj = j + k <= n ? _rank[j + k] : -1;
		return ri < rj;
	}
}
void construct_sa(string S, int* sa) {
	n = S.length();

	for (size_t i = 0; i <= n; i++)
	{
		sa[i] = i;
		_rank[i] = i < n ? S[i] : -1;
	}

	for ( k = 1; k <= n; k *= 2)
	{
		sort(sa, sa + n + 1, compare_sa);

		// saはソート後の接尾辞の並びになっている、rankは元の位置のindexを保持したまま、更新されている。
		// ピンとこなかった部分
		temp[sa[0]] = 0;
		for (size_t i = 1; i <= n; i++)
		{
			temp[sa[i]] = temp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
		}
		for (size_t i = 0; i <= n; i++)
		{
			_rank[i] = temp[i];
		}
	}
}
bool contain(string S, int* sa, string T) {
	int a = 0, b = S.length();
	// sa は 接尾辞が辞書順に並んでいる、入っているのはその位置のインデックス
	while (b - a > 1) {
		int c = (a + b) / 2;
		if (S.compare(sa[c], T.length(), T) < 0) a = c;
		else b = c;
	}
	return S.compare(sa[b], T.length(), T) == 0;
}


#define bit(x,v) ((ll)x << v)

class BIT {

	static const int MAX_N = 400010;
public:
	BIT() { memset(bit, 0, sizeof(bit)); }
	ll bit[MAX_N + 1], n;
	ll sum(int i) {
		ll s = 0;
		while (i > 0)
		{
			s += bit[i];
			i -= i & -i;
		}
		return s;
	}

	void add(int i, int x) {
		while (i <= n)
		{
			bit[i] += x;
			i += i & -i;
		}
	}


}; 

double to_deg(double r) {
	return r * 180.0 / (atan(1.0) * 4.0);
}

void solv() {

	ll k;
	cin >> n >> k; 
	ll a[200010],f[200010];
	ll sum = 0;
	for (size_t i = 0; i < n; i++)
	{
		cin >> a[i];
		sum += a[i];
	}

	for (size_t i = 0; i < n; i++)
	{
		cin >> f[i];
	}
	if (sum <= k) {
		cout << 0 << endl;
		return;
	}

	sort(a, a + n);
	sort(f, f + n);
	reverse(f, f + n);
	ll p[200010];
	ll ma = 0;
	for (size_t i = 0; i < n; i++)
	{
		p[i] = a[i] * f[i];
		ma = max(ma, p[i]);
	}

	ll hi = ma;
	ll lo = 0;
	int i = 0;
	ll mid = 0;
	ll r = INF * INF;
	while (i < 100)
	{
		mid = (hi + lo) / 2;
		ll kc = 0;
		ll pr = 0;
		for (size_t i = 0; i < n; i++)
		{
			if (mid < p[i]) {
				ll v = (mid - p[i]) ;
				if (v % f[i] != 0) {
					kc++;
				}
				
				kc += v / (-f[i]) ;
				pr = max(p[i] / mid,pr);
				if (kc > k) {
					break;
				}
			}
			else pr = max(pr, p[i]);
		}
		if (kc > k) {
			lo = mid;
		}
		else {
			r = min(pr, r);
			hi = mid;
		}
		i++;
	}

	cout << r << endl;

	//double a, b, x;
	//cin >> a >> b >> x;

	//double px = (x / a) / a;
	//double pd = (b - px) * 2;
	//double c2 = a * a + pd * pd;
	//double c = sqrt(c2);
	//double cs = (a * a + c2 - (pd * pd)) / (2.0 * a * c);
	//double rs = to_deg( atan(pd / a));
	//double p = 1.0 / sqrt(2);
	//cout << rs << endl;
}

int main() {
	//COMinit();
	solv();
	return 0;
}