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.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int suma=0;
		int sumb=0;
		int kk=0;
		Vx[] c=new Vx[n+1];
		c[0]=new Vx(9999,9999);
		for(int i=1;i<=n;i++)
		{
			int t=sc.nextInt();
			c[i]=new Vx(t,0);
			suma+=t;
		}
		
		for(int i=1;i<=n;i++)
		{
			
			c[i].b=sc.nextInt();
		}
		Arrays.sort(c);
		while(sumb<suma)
		{
			sumb+=c[++kk].b;
		}
		int[][] dp=new int[10005][105];
		for(int i=0;i<10005;i++)Arrays.fill(dp[i],-9999);
		dp[0][0]=0;
		for(int i=1;i<=n;i++)
		{
			for(int j=sumb;j>=c[i].b;j--)
			{
				for(int k=1;k<=kk;k++)
				{
					dp[j][k]=Math.max(dp[j][k],dp[j-c[i].b][k-1]+c[i].a);
				}
			}
		}
		int ans=0;
		for(int i=suma;i<=sumb;i++)ans=Math.max(ans, dp[i][kk]);
		System.out.print(kk+" "+(suma-ans));
	}

}
class Vx implements Comparable<Vx>
{
	int a;
	int b;
	Vx(){}
	Vx(int a,int b){
		this.a=a;
		this.b=b;
	}
	@Override
	public int compareTo(Vx o) {
		// TODO 自动生成的方法存根
		return o.b-this.b;
	}
	
	
	
}