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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)

using namespace std;
typedef pair<int, int> P;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

//最大公約数
ll gcd(ll x, ll y) {
    ll tmp = 0;
    if (x < y){
        tmp=x;
        x=y;
        y=tmp;
    }
    while (y > 0) {
        ll r = x % y;
        x = y;
        y = r;
        }
    return x;
}

//最大公倍数
ll lcm(ll x,ll y){
    return x/gcd(x,y)*y;
}

const int MAX = 1e6+1;
const ll MOD = 998244353;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
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;
    }
}

struct mint {
  ll x; // typedef long long ll;
    mint(ll x=0):x((x%MOD+MOD)%MOD){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
    if ((x += a.x) >= MOD) x -= MOD;
    return *this;
    }
    mint& operator-=(const mint a) {
    if ((x += MOD-a.x) >= MOD) x -= MOD;
    return *this;
    }
    mint& operator*=(const mint a) {
        (x *= a.x) %= MOD;
        return *this;
    }
    mint operator+(const mint a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }

    // for prime MOD
    mint inv() const {
        return pow(MOD-2);
    }
    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint a) const {
        mint res(*this);
        return res/=a;
    }
};

// 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;
}


//階乗
ll kaijo(ll k){
    ll sum = 1;
    for (ll i = 1; i <= k; ++i)
    {
        sum *= i;
        sum%=1000000000+7;
    }
    return sum;
}

long long modpow(long long a, long long n, long long MOD) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % MOD;
        a = a * a % MOD;
        n >>= 1;
    }
    return res;
}

//sieve(MAX);でベクトルIsPrimeに「MAXまでの数値の素数の真偽」を格納する。
struct Sieve {
    int n;
    vector<int> f, primes;
    Sieve(int n=1):n(n), f(n+1) {
        f[0] = f[1] = -1;
        for (ll i = 2; i <= n; ++i) {
        if (f[i]) continue;
        primes.push_back(i);
        f[i] = i;
        for (ll j = i*i; j <= n; j += i) {
            if (!f[j]) f[j] = i;
        }
        }
    }
    bool isPrime(int x) { return f[x] == x;}
    vector<int> factorList(int x) {
        vector<int> res;
        while (x != 1) {
        res.push_back(f[x]);
        x /= f[x];
        }
        return res;
    }
    vector<P> factor(int x) {
        vector<int> fl = factorList(x);
        if (fl.size() == 0) return {};
        vector<P> res(1, P(fl[0], 0));
        for (int p : fl) {
        if (res.back().first == p) {
            res.back().second++;
        } else {
            res.emplace_back(p, 1);
        }
        }
        return res;
    }
};
struct UnionFind {
    vector<int> par;
    UnionFind(int n) : par(n, -1) { }
    int root(int x) {
        if (par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }
    bool issame(int x, int y) {
        return root(x) == root(y);
    }
    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (par[x] > par[y]) swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }
    int size(int x) {
        return -par[root(x)];
    }
};

ll count(int n, int a){
    ll bunshi = 1;
    for(int i = 0; i < a; i++){
        bunshi*=(n-i);
        bunshi%=MOD;
    }
    ll bunbo = 1;
    for(int i = 1; i < a+1; i++){
        bunbo*=i;
        bunbo%=MOD;
    }
    bunbo = modpow(bunbo,MOD-2,MOD);
    // cout<<(bunshi*bunbo)%MOD<<endl;
    return (bunshi*bunbo)%MOD;
}

// 約数列挙。約数をvector<ll>で返す。計算量はsqrt(n)。
vector<ll> divisor(ll n) {
    vector<ll> ret;
    for(ll i = 1; i * i <= n; i++) {
        if(n % i == 0) {
        ret.push_back(i);
        if(i * i != n) ret.push_back(n / i);
        }
    }
    sort(begin(ret), end(ret));
    return (ret);
}

// ここから開始
const ll INF = 1e18;
int dx[] = {1,0,-1,0};
int dy[] = {0,1,0,-1};

int main(){
    int N,K;
    cin>>N>>K;
    vector<int> a(N);
    vector<int> f(N);
    rep(i,N) cin>>a[i];
    rep(i,N) cin>>f[i];
    sort(f.begin(),f.end());
    reverse(f.begin(),f.end());
    sort(a.begin(),a.end());

    // 二分探索
    ll left = -1, right = 1e12;
    while(right-left>1){
        ll mid = (left+right)/2;
        ll tmp = 0;
        rep(i,N){
            tmp += max((ll)0, a[i]-mid/f[i]);
        }
        if(tmp<=K) right = mid;
        else left = mid;
    }
    cout<<right<<endl;
    // cout<<ans<<endl;
    return 0;
}