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: Accepted)

#include <iostream>
#include <stdio.h>
#include <cmath>
#include <climits>
#include <algorithm>
#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <string.h>
#include <iomanip>
#include <ctype.h>

#define ll          long long int
#define ull         unsigned long long int
#define pb          push_back
#define pii         pair<int,int>
#define F         first
#define S          second
#define vi          vector<int>
#define vli          vector<ll>
#define vii         vector<pii>
#define mi          map<int,int>
#define mii         map<pii,int>
#define M_PI        3.14159265358979323846
#define mp          make_pair
#define sz(x)       (int)x.size()
#define hell        1000000007
#define forn(i,a,b)  for(int i=a;i<b;i++)
#define eps 1e9
#define all(a)      (a).begin(),(a).end()
const long long INF = (long long) 1e12+1;


using namespace std;
void solve();

void fastIO() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}

int main(){
   fastIO();
    solve();
}
//void sieve();
ll pw(ll k,ll x);
ll gcd(ll x,ll y);


/*class CompareDist
{
public:
    bool operator()(pair<ll,int> a,pair<ll,int> b) {
        if(a.F!=b.F)
        return a.F<b.F;
        else
        return a.S>b.S;
    }
};*/
int n;
ll k,a[1000001],f[1000001];
bool F(ll x);

void solve(){

  cin>>n>>k;
  forn(i,0,n)
  cin>>a[i];
  forn(i,0,n)
  cin>>f[i];

  sort(a,a+n);
  sort(f,f+n);

  ll lo=0,hi=INF;
  ll ans;
  while(lo<=hi){
    ll mid=lo+(hi-lo)/2;
    if(F(mid))
    {
      ans=mid;
      hi=mid-1;
    }
    else{
      lo=mid+1;
    }
  }
  cout<<ans;
}

bool F(ll x){
  ll sum=0;
  for(int i=0;i<n;i++){
    ll y=x/f[i];
    sum+=max(0ll,a[n-i-1]-y);
  }
  if(sum<=k)
  return true;
  else
  return false;
}

///Before submit=>
///    *check for integer overflow,array bounds
///    *check for n=1

ll gcd(ll x,ll y){
  if(x==0)
  return y;
  else
  return gcd(y%x,x);
}


ll pw(ll k,ll x){
  if(x==0)
  return 1;
  if(x&1)
  return ((pw(k,x-1)%hell)*k%hell)%hell;
  else
  return ((pw(k,x/2)%hell)*(pw(k,x/2)%hell))%hell;
}

/*void sieve(){

  for(int i=2;i<=1000000;i++){
    if(prime[i]==true){
      for(int j=i*i;j<=1000000;j+=i)
      prime[j]=false;
    }
  }

}*/