Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai ≤ bi).

Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.

Nick asks you to help him to determine k — the minimal number of bottles to store all remaining soda and t — the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.

Input

The first line contains positive integer n (1 ≤ n ≤ 100) — the number of bottles.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the amount of soda remaining in the i-th bottle.

The third line contains n positive integers b1, b2, ..., bn (1 ≤ bi ≤ 100), where bi is the volume of the i-th bottle.

It is guaranteed that ai ≤ bi for any i.

Output

The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.

Examples

Input

4
3 3 4 3
4 7 6 5


Output

2 6


Input

2
1 1
100 100


Output

1 1


Input

5
10 30 5 6 24
10 41 7 8 24


Output

3 11

Note

In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Bottles {
	static int n,a[],b[];
	static Pair memo[][];
	public static Pair dp(int index,int curVol)
	{
		if (index==n)
		{
			if (curVol>=0)
				return new Pair(0,0);
			else
				return new Pair(10000,0);
		}
		if (memo[index][curVol+10000]!=null)
			return memo[index][curVol+10000];
		Pair p1 = dp(index+1,curVol+b[index]-a[index]);
		Pair p2 = dp(index+1,curVol-a[index]);
		Pair res;
		if (p1.k+1<p2.k)
			res =  new Pair(p1.k+1,p1.t);
		else if (p1.k+1>p2.k)
			res = new Pair(p2.k,p2.t+a[index]);
		else
			res = new Pair(p2.k,Math.min(p1.t, p2.t+a[index]));
		return memo[index][curVol+10000]= res;
	}
	public static void main(String args[]) throws IOException
	{
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		a = new int[n];
		b = new int[n];
		memo = new Pair[n][20005];
		for (int i=0;i<n;i++)
			a[i] = sc.nextInt();
		for (int i=0;i<n;i++)
			b[i] = sc.nextInt();
		System.out.println(dp(0,0));
	}
	static class Pair
	{
		int k;
		int t;
		public Pair(int k,int t)
		{
			this.k = k;
			this.t = t;
		}
		public String toString()
		{
			return k+" "+t;
		}
	}
	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner(InputStream is) {
			br = new BufferedReader(new InputStreamReader(is));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		long nextLong() throws IOException {
			return Long.parseLong(next());
		}
	}
}
