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.*;
import java.util.*;
public class no {
	static int n,c,total;
	static int [] a,b;
	public static void main(String[] args) throws Throwable {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw=new PrintWriter(System.out,true);

		n=Integer.parseInt(br.readLine());
		a=new int [n];
		b=new int [n];
		String s = br.readLine();
		String[] sa = s.split(" ");
		for(int i=0;i<n;i++){
			a[i] = Integer.parseInt(sa[i]);
			total+=a[i];
		}
		s = br.readLine();
		sa = s.split(" ");
		for(int i=0;i<n;i++){
			b[i] = Integer.parseInt(sa[i]);
		}
		pair[] arr = new pair[n];
		for(int i = 0;i<n;i++)
			arr[i] = new pair(a[i],b[i]);

		Arrays.sort(arr);

		int j=0;
		int curSum = 0;
		while(j<n && curSum<total)
		{
			curSum+=arr[j].b;
			j++;
		}
		c= j;

		mem=new int [n+1][(int)1e4+1][c+1];
		for(int i=0;i<=n;i++)
			for(int k=0;k<mem[i].length;k++)
				Arrays.fill(mem[i][k], -1);
		
		//System.out.println(dp(0, total,c));
		pw.println(c + " " +(total-dp(0, total,c)));

		pw.flush();
		pw.close();
	}
	static int [][][] mem;
	static int inf=(int)(1e7);
	static int dp(int i,int v , int remC){
//		System.out.println(i + " " + v + " " + remC);
		if(v<=0 && remC==0)
			return 0;
		if(v<0 || remC<0)
			return -inf;
		if(i == n)
			return -inf;
		if(mem[i][v][remC]!=-1)
			return mem[i][v][remC];

		int ans=dp(i+1, v,remC);
		ans=Math.max(ans, a[i]+dp(i+1, v-b[i],remC-1));

		return mem[i][v][remC]=ans;
	}

	static class pair implements Comparable<pair>{
		int b,a;

		pair(int a,int b){
			this.a=a;
			this.b=b;
		}
		public int compareTo(pair o) {
			if(b!=o.b)
				return o.b-b;
			return o.a-a;
		}
	}
}
