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 <bits/stdc++.h>
using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}

//typedef
//------------------------------------------
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL,LL> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;
typedef vector<bool> VB;
typedef vector<VB> VVB;
typedef vector<double> VD;
typedef vector<VD> VVD;
typedef vector<string> VS;
typedef vector<VS> VVS;
typedef vector<char> VC;
typedef vector<VC> VVC;
typedef vector<PII> VPII;
typedef vector<PLL> VPLL;
typedef priority_queue<int> PQGI; //大きい順
typedef priority_queue<int, VI, greater<int>> PQLI;
typedef priority_queue<PII> PQGP;
typedef priority_queue<PII, VPII, greater<int>> PQLP;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define EB emplace_back
#define PB push_back
#define PF push_front
#define POB pop_back
#define POF pop_front
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a)*(a))
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define SORTR(c) sort((c).rbegin(), (c).rend())
#define LB lower_bound
#define UB upper_bound
#define NEXP next_permutation
#define FI first
#define SE second
#define Vmin(a) *min_element((a).begin(),(a).end())
#define Vmax(a) *max_element((a).begin(),(a).end())

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define FORR(i,a,b) for(int i = (b-1);i>=a;i--)
#define REPR(i,n) FORR(i,0,n)
#define BFOR(bit,a,b) for(int bit = (a); bit < (1<<(b)); ++bit)
#define BREP(bit,n) BFOR(bit,0,n)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
const int INF = INT_MAX/2;
const LL LINF = LLONG_MAX/3;
const int RINF = INT_MIN/2;
const LL RLINF = LLONG_MIN/3;
const LL MOD = 1e9+7;
const LL MODD = 998244353;
const int MAX = 510000;

inline bool Eq(double a, double b) { return fabs(b - a) < EPS; }

//other
//-------------------------------------------
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

#define COUT(x) cout << (x) << endl
#define COUT2(x,y) cout << (x) << " " << (y) << endl
#define COUT3(x,y,z) cout << (x) << " " << (y) << " " << (z) << endl
#define PR(x) cout << (x)
#define ENDL cout << endl
#define SPACE cout << " "
#define BC(x) __builtin_popcountll(x)


VI dx = {1,0,-1,0,1,1,-1,-1};
VI dy = {0,1,0,-1,1,-1,1,-1};
VC dc = {'R','D','L','U'};
VS ds = {"R","D","L","U","RD","RU","LD","LU"};

LL Gcd(LL a, LL b) {return __gcd(a,b);}
LL Lcm(LL a, LL b) {return a/Gcd(a,b)*b;}

LL ModPow(LL A, LL N, LL M){
  LL res = 1;
  while(N>0){
    if(N&1) res = res * A % M;
    A = A * A % M;
    N >>= 1;
  }
  return res;
}


template<class T> void VCout(vector<T> &v){
  int N = SZ(v);
  REP(i,N){
    cout << v.at(i);
    if(i == N-1) cout << endl;
    else cout << " ";
  }
  return;
}

VLL a, f;
int n;
LL k;

// index が条件を満たすかどうか
bool isOK(LL key) {
    LL res=0;
    REP(i,n){
      if(a.at(i)*f.at(i) <= key) continue;
      res += a.at(i) - key / f.at(i);
    }
    return res<=k;
}

// 汎用的な二分探索のテンプレ
int binary_search() {
    LL ok = 1e13; //「index = 0」が条件を満たすこともあるので、初期値は -1
    LL ng = -1; // 「index = a.size()-1」が条件を満たさないこともあるので、初期値は a.size()

    /* どんな二分探索でもここの書き方を変えずにできる！ */
    while (abs(ok - ng) > 1) {
        LL mid = (ok + ng) / 2;

        if (isOK(mid)) ok = mid;
        else ng = mid;
    }
    return ok;
}



int main() {

  // cin.tie(0);
  // ios::sync_with_stdio(false);
  cout << fixed << setprecision(12);

  cin >> n;
  cin >> k;
  a.assign(n,0);
  f.assign(n,0);
  REP(i,n) cin >> a.at(i);
  REP(i,n) cin >> f.at(i);
  SORT(a);
  SORTR(f);
  
  COUT(binary_search());
  
  return 0;
}