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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*; import java.io.*; import java.math.*;
public class Main{
	//Don't have to see. start------------------------------------------
	static class InputIterator{
		ArrayList<String> inputLine = new ArrayList<String>(1024);
		int index = 0; int max; String read;
		InputIterator(){
			BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
			try{
				while((read = br.readLine()) != null){
					inputLine.add(read);
				}
			}catch(IOException e){}
			max = inputLine.size();
		}
		boolean hasNext(){return (index < max);}
		String next(){
			if(hasNext()){
				return inputLine.get(index++);
			}else{
				throw new IndexOutOfBoundsException("There is no more input");
			}
		}
	}
	static InputIterator ii = new InputIterator();//This class cannot be used in reactive problem.
	static PrintWriter out = new PrintWriter(System.out);
	static void flush(){out.flush();}
	static void myout(Object t){out.println(t);}
	static void myerr(Object t){System.err.print("debug:");System.err.println(t);}
	static String next(){return ii.next();}
	static boolean hasNext(){return ii.hasNext();}
	static int nextInt(){return Integer.parseInt(next());}
	static long nextLong(){return Long.parseLong(next());}
	static ArrayList<String> nextStrArray(){return myconv(next(), 8);}
	static ArrayList<String> nextCharArray(){return myconv(next(), 0);}
	static ArrayList<Integer> nextIntArray(){
		ArrayList<String> input = nextStrArray(); ArrayList<Integer> ret = new ArrayList<Integer>(input.size());
		for(int i = 0; i < input.size(); i++){
			ret.add(Integer.parseInt(input.get(i)));
		}
		return ret;
	}
	static ArrayList<Long> nextLongArray(){
		ArrayList<String> input = nextStrArray(); ArrayList<Long> ret = new ArrayList<Long>(input.size());
		for(int i = 0; i < input.size(); i++){
			ret.add(Long.parseLong(input.get(i)));
		}
		return ret;
	}
	static HashMap<Integer, String> CONVSTR = new HashMap<Integer, String>();
	static String myconv(Object list, int no){//only join
		String joinString = CONVSTR.get(no);
		if(list instanceof String[]){
			return String.join(joinString, (String[])list);
		}else if(list instanceof ArrayList){
			return String.join(joinString, (ArrayList)list);
		}else{
			throw new ClassCastException("Don't join");
		}
	}
	static ArrayList<String> myconv(String str, int no){//only split
		String splitString = CONVSTR.get(no);
		return new ArrayList<String>(Arrays.asList(str.split(splitString)));
	}
	public static void main(String[] args){
		CONVSTR.put(8, " "); CONVSTR.put(9, "\n"); CONVSTR.put(0, "");
		Runtime rt = Runtime.getRuntime();
		long mae = System.currentTimeMillis();
		solve();flush();
		long ato = System.currentTimeMillis();
		myerr("time : " + (ato - mae) + "ms");
		myerr("memory : " + ((rt.totalMemory() - rt.freeMemory()) / 1024) + "KB");
	}
	//Don't have to see. end------------------------------------------
	static void solve(){//Here is the main function
		ArrayList<Long> one = nextLongArray();
		int N = Integer.parseInt(String.valueOf(one.get(0)));
		long K = one.get(1);
		ArrayList<Long> alist = nextLongArray();
		ArrayList<Long> flist = nextLongArray();
		Collections.sort(alist);
		Collections.sort(flist, Comparator.reverseOrder());
		long L = -1;
		long R = Long.parseLong("20000000000009");
		while(R - L > 1){
			long C = (L + R) / 2;
			long tmpK = K;
			boolean isOK = true;
			for(int i = 0; i < N; i++){
				long now = alist.get(i) * flist.get(i);
				long req = now - C;
				if(req > 0){
					tmpK -= (req + flist.get(i) - 1) / flist.get(i);
					if(tmpK < 0){
						isOK = false;
						break;
					}
				}
			}
			if(isOK){
				R = C;
			}else{
				L = C;
			}
		}
		
		myout(Math.max(0, R));
	}
	//Method addition frame start

	//Method addition frame end
}