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 J {
	
	public static void main(String[] args) {
		
		Scanner qwe= new Scanner(System.in);
		
		int N = qwe.nextInt();
		
		int[] h = new int[N];
		for (int i = 0; i < h.length; i++) {
			h[i] = qwe.nextInt();
		}
		int[] c = new int[N];
		for (int i = 0; i < c.length; i++) {
			c[i] = qwe.nextInt();
		}
		
		int[][][] dp = new int[2][N+1][10002];
		
		int toadd = 5000;
		
		int oo = 987654321;
		
		//base case
		
		for(int x = 0; x < dp[0][0].length; x++){
			dp[1][0][x] = (x >= toadd) ? 0 : oo;
			for(int k = 1; k <= N; k++) dp[1][k][x] = oo;
		}
		
		for(int i = 0; i < N; i++){
			int prev = 1 - (i & 1);
			
			for(int k = 0; k <= N; k++){
				for(int x = 0; x < dp[0][0].length; x++){
					dp[i&1][k][x] = oo;
					
					if(k > 0){
						int takex = Math.min(x + c[i]-h[i],dp[0][0].length-1);
						int takek = k-1;
						
						dp[i&1][k][x] = dp[prev][takek][takex];
						
					}
					
					int dontx = x-h[i];
					if(dontx >= 0){
						dp[i&1][k][x] = Math.min(dp[i&1][k][x], dp[prev][k][dontx]+h[i]);
					}
					
				}
			}
			
			
		}
		
		int ansk = -1;
		int anscost = -1;
		
		for(int k = N; k >= 1; k--){
			int bestans = oo;
				bestans = Math.min(bestans, dp[(N-1)&1][k][toadd]);
			if(bestans < oo){
				ansk = k;
				anscost = bestans;
			}
		}
		
		System.out.println(ansk + " " + anscost);
		
		qwe.close();
	}

}
