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.*;
import java.math.*;
public class main{
	
	public static void main(String args[])
	{
		int a[]=new int[101];
		int b[]=new int[101];
		int d[]=new int[10010];
		int v[]=new int[10010];
		Scanner cin=new Scanner(System.in);
		int n=cin.nextInt();
		int sum=0;
		for(int i=0;i<n;i++)
		{
			a[i]=cin.nextInt();
			sum+=a[i];
		}
		for(int i=1;i<=sum;i++)d[i]=1000000007;
		for(int i=0;i<n;i++)b[i]=cin.nextInt();
		for(int i=0;i<n;i++)
		{
			for(int j=sum;j>=0;j--)
			{
				if(d[Math.min(j+b[i],sum)]>d[j]+1)
				{
					d[Math.min(j+b[i],sum)]=d[j]+1;
					v[Math.min(j+b[i],sum)]=v[j]+a[i];
				}
				else if(d[Math.min(j+b[i],sum)]==d[j]+1&&v[Math.min(j+b[i],sum)]<v[j]+a[i])
				{
					v[Math.min(j+b[i],sum)]=v[j]+a[i];
				}
			}
		}
		System.out.print(d[sum]+" "+(sum-v[sum]));
	}
}