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.Arrays;
import java.util.HashMap;
import java.util.Scanner;

public class J {
	static final int inf = (int)1e7;
	static int n;
	static int[] as,bs;
	static int[][][] memo;
	static final int maxa = 100*100;
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		n = in.nextInt();
		as = new int[n];
		bs = new int[n];
		int totalA = 0;
		for(int i=0;i<n;i++) {
			as[i] = in.nextInt();
			totalA += as[i];
		}
		for(int i=0;i<n;i++)
			bs[i] = in.nextInt();
		int[] bsSorted = bs.clone();
		Arrays.sort(bsSorted);
		
		int mink = 0;
		
		int curA = 0;
		for(int j=bsSorted.length-1;j>=0;j--) {
			if(curA >= totalA) break;
			curA += bsSorted[j];
			mink++;
		}
		memo = new int[2][(mink+1)][maxa+1];
		for(int[][] arr2 : memo)
			for(int[] arr1 : arr2)
				Arrays.fill(arr1, inf);
		
		int par = 0;
		int minAns = inf;
		memo[par][0][0] = 0;
		for(int idx=0;idx<n;idx++) {
			par = 1-par;
			
			for (int ku = 0; ku <= mink; ku++) {
				Arrays.fill(memo[par][ku], inf);
				for(int a=0;a<=maxa;a++) {
					memo[par][ku][a] = Math.min(memo[par][ku][a], memo[1-par][ku][a]+as[idx]);
					if(ku > 0 && a-bs[idx] >= 0)
						memo[par][ku][a] = Math.min(memo[par][ku][a], memo[1-par][ku-1][a-bs[idx]]);
					if(idx == n-1 && a >= totalA && ku == mink)
						minAns = Math.min(minAns, memo[par][ku][a]);
				}
//				System.out.println("memo[" + idx + "][" + ku + "] = " + Arrays.toString(memo[par][ku]));
			}
		}
		System.out.println(mink + " " + minAns);
	}
}
