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



typedef unsigned long ul;
typedef unsigned long long ull;
typedef long long ll;
typedef vector<ll> vint;
typedef vector<vector<ll>> vvint;
typedef vector<vector<vector<ll>>> vvvint;
typedef vector<string> vstring;
typedef vector<vector<string>> vvstring;
typedef vector<char> vchar;
typedef vector<vector<char>> vvchar;
typedef vector<long double> vdouble;
typedef vector<vector<long double>> vvdouble;
typedef vector<vector<vector<long double>>> vvvdouble;
typedef pair<ll,ll> pint;
typedef vector<pint> vpint;
typedef vector<bool> vbool;

#define rep(i,n) for(ll i=0;i<n;i++)
#define repf(i,f,n) for(ll i=f;i<n;i++)
#define repr(i,n) for(ll i=n-1;i>=0;i--)
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define ALL(obj) (obj).begin(), (obj).end()
// #define LLONG_MAX 9223372036854775806
#define vmax(vec) *max_element(vec.begin(), vec.end())
#define vmin(vec) *min_element(vec.begin(), vec.end())
#define vsort(vec) sort(vec.begin(), vec.end())
#define vsortgr(vec) sort(vec.begin(), vec.end(), greater<ll>())
#define MOD 1000000007
// #define MOD 998244353

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; }

//
// struct Node{
//   vint children;
//   ll index;
//   ll prop;
// };
// struct edge{ll to; ll cost;};



int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0,};

template< typename T >
struct edge {
  int src, to;
  T cost;

  edge(int to, T cost) : src(-1), to(to), cost(cost) {}

  edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}

  edge &operator=(const int &x) {
    to = x;
    return *this;
  }

  operator int() const { return to; }
};

template< typename T >
using Edges = vector< edge< T > >;
template< typename T >
using WeightedGraph = vector< Edges< T > >;
using UnWeightedGraph = vector< vector< int > >;
template< typename T >
using Matrix = vector< vector< T > >;

ll gcd(ll a, ll b)
{
    if (a < b) {
        a ^= b;
        b ^= a;
        a ^= b;
    }

    return b ? gcd(b, a % b) : a;
}

ll lcm(int a, ll b)
{
    return a * b / gcd(a, b);
}




// 繰り返し二乗法
ll power(ll a, ll b){
	ll res=1;
	while(b>0){
		if(b&1) res=res*a%MOD;
		a=a*a%MOD;
		b>>=1;
	}
	return res;
}

const int MAX=510000;

ll fact[MAX], fact_inv[MAX];

void init_fact(ll n){
  fact[0]=1;
  // 階乗の計算
  rep(i, n) fact[i+1]=fact[i]*(i+1)%MOD;
  fact_inv[n]=power(fact[n], MOD-2);
  // 逆元の計算
  for(ll i=n-1;i>=0;i--) fact_inv[i]=fact_inv[i+1]*(i+1)%MOD;
}

ll comb(ll n, ll r){
	return (fact[n]*fact_inv[r])%MOD*fact_inv[n-r]%MOD;
}

ll perm(ll n, ll r){
  return (fact[n]*fact_inv[n-r])%MOD;
}


struct UnionFind {
    vector<ll> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
    vector<ll> num;
    // vector<ll> dist; // rootまでの距離

    UnionFind(ll N) : par(N), num(N){ //最初は全てが根であるとして初期化
      for(ll i = 0; i < N; i++) par[i] = i;
      for(ll i = 0; i < N; i++) num[i] = 1;
      // for(ll i = 0; i < N; i++) dist[i] = 0;

    }

    ll root(ll x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(ll x, ll y) { // xとyの木を併合
        ll rx = root(x); //xの根をrx
        ll ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
        num[ry] = num[rx] + num[ry];
    }



    bool same(ll x, ll y) { // 2つのデータx, yが属する木が同じならtrueを返す
        ll rx = root(x);
        ll ry = root(y);
        return rx == ry;
    }

    ll size(ll x){
      return num[root(x)];
    }


};

vint divisor(ll n){ // nの約数
  vint ret;
  for(ll i=1;i*i<=n;i++){
    if(n%i == 0){
      ret.pb(i);
      if(i*i!=n) ret.pb(n/i);
    }
  }
  vsort(ret);
  return ret;
}

ll my_pow(ll x,ll y){
  ll rtn=1;
  rep(i,y) rtn*=x;
  return rtn;
}

ll get_digit_in(ll n, ll i){ // i桁目の数字を得る。
  for(ll j=0;j<i-1;j++){
    n/=10;
  }
  return n%10;
}

ll get_digit(ll n){
  ll rtn=0;
  while(n>0){
    n/=10;
    rtn++;
  }
  return rtn;
}

map<ll,ll> prime_factor(ll n){
  map<ll,ll> rtn;
  for(ll i=2;i*i<=n;i++){
    while(n%i == 0){
      rtn[i]++;
      n /= i;
    }
  }
  if(n!=1) rtn[n] = 1;
  return rtn;
}

// nPnの順列に対して処理を実行する
void foreach_permutation(ll n, function<void(ll *)> f) {
  ll indexes[n];
  for (ll i = 0; i < n; i++) indexes[i] = i;
  do {
    f(indexes);
  } while (std::next_permutation(indexes, indexes + n));
}

void recursive_comb(ll *indexes, ll s, ll rest, function<void(ll *)> f) {
  if (rest == 0) {
    f(indexes);
  } else {
    if (s < 0) return;
    recursive_comb(indexes, s - 1, rest, f);
    indexes[rest - 1] = s;
    recursive_comb(indexes, s - 1, rest - 1, f);
  }
}

// nCkの組み合わせに対して処理を実行する
void foreach_comb(ll n, ll k, function<void(ll *)> f) {
  ll indexes[k];
  recursive_comb(indexes, n - 1, k, f);
}

// nPkの順列に対して処理を実行する
void foreach_permutation(ll n, ll k, function<void(ll *)> f) {
  foreach_comb(n, k, [&](ll *c_indexes) {
    foreach_permutation(k, [&](ll *p_indexes) {
      ll indexes[k];
      for (ll i = 0; i < k; i++) {
        indexes[i] = c_indexes[p_indexes[i]];
      }
      f(indexes);
    });
  });
}

// ll arr[1000500];
// vint primes;

// //エラトステネスの篩
// void Eratosthenes(){
//   ll N = 1000500;
// 	for(ll i = 0; i < N; i++){
// 		arr[i] = 1;
// 	}
// 	for(ll i = 2; i < sqrt(N); i++){
// 		if(arr[i]){
// 			for(ll j = 0; i * (j + 2) < N; j++){
// 				arr[i *(j + 2)] = 0;
// 			}
// 		}
// 	}

// 	for(ll i = 2; i < N; i++){
// 		if(arr[i]){
//       primes.pb(i);
// 			//cout << i << endl;
// 		}
// 	}
// }


void print(vint &V){
  for(auto e:V)cout<<e<<" ";
  cout<<endl;
}


ll solve(const ll K, const vint &A, const vint &F){
  ll left = -1;
  ll right = 1000000000000;

  while(right-left > 1){
    ll mid = left + (right - left)/2;
    // cout<<mid<<endl;

    bool f = true;
    ll cnt = 0;
    rep(i,A.size()){
      if(A[i]*F[i] > mid){
        cnt += A[i]-mid/F[i];
      }
      if(cnt > K){
        f = false;
        break;
      }
    }

    if(f) right=mid;
    else left=mid;


  }

  return right;
}



int main() {
  cout<<fixed<<setprecision(12);


  ll N,K;cin>>N>>K;
  vint A(N);rep(i,N)cin>>A[i];
  vint F(N);rep(i,N)cin>>F[i];


  sort(ALL(A));
  sort(ALL(F), greater<ll>());

  ll ans = solve(K,A,F);

  cout<<ans<<endl;





}