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.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Solveaproblem {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(reader.readLine());
        StringTokenizer sToken = new StringTokenizer(reader.readLine());
        int[] a = new int[n];
        int sum = 0;
        for (int i=0; i<n; i++) {
            a[i] = Integer.parseInt(sToken.nextToken());
            sum+=a[i];
        }
        sToken = new StringTokenizer(reader.readLine());
        int[] c = new int[n];
        for (int i=0; i<n; i++) c[i] = Integer.parseInt(sToken.nextToken()) - a[i];
        int[][] dp = new int[n+1][sum+1];
        int min = -10000000;
        for (int i=0; i<=n; i++) {
            for (int j=0; j<=sum; j++) {
                dp[i][j] = min;
            }
        }
        dp[0][0] = 0;
        for (int i=0; i<n; i++) {
            int[][] newdp = new int[n+1][sum+1];
            for (int k=0; k<=sum; k++) {
                if (k>=a[i]) newdp[0][k] = dp[0][k-a[i]];
                else newdp[0][k] = min;
            }
            for (int j=1; j<=n; j++) {
                for (int k=0; k<a[i]; k++) {
                    if (dp[j-1][k]!=min) newdp[j][k] = dp[j-1][k] + c[i];
                    else newdp[j][k] = min;
                }
                for (int k=a[i]; k<=sum; k++) {
                    if (dp[j-1][k]==min) newdp[j][k] = dp[j][k-a[i]];
                    else newdp[j][k] = Math.max(dp[j-1][k] + c[i], dp[j][k-a[i]]);
                }
            }
            for (int j=0; j<=n; j++) {
                for (int k=0; k<=sum; k++) {
                    dp[j][k] = newdp[j][k];
                }
            }
        }
        for (int i=0; i<=n; i++) {
            for (int j=0; j<=sum; j++) {
                if (dp[i][j]>=j) {
                    PrintWriter writer = new PrintWriter(System.out);
                    writer.print(i+" "+j);
                    writer.close();
                    System.exit(0);
                }
            }
        }
    }
}