2104.Bottles

Time Limit: 1s Memory Limit: 256MB

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 \le 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 Format(From the terminal/stdin)

The first line contains positive integer n (1 \le n \le 100) - the number of bottles.

The second line contains n positive integers a1,a2,...,an (1 \le ai \le 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 \le bi \le 100), where bi is the volume of the i-th bottle.

It is guaranteed that ai \le bi for any i.

Output Format(To the terminal/stdout)

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.

Sample Input 1

Copy
4
3 3 4 3
4 7 6 5
 \n
 · · · \n
 · · · \n

Sample Output 1

Copy
2 6
 · \n

Sample Input 2

Copy
2
1 1
100 100
 \n
 · \n
   ·   \n

Sample Output 2

Copy
1 1
 · \n

Sample Input 3

Copy
5
10 30 5 6 24
10 41 7 8 24
 \n
  ·  · · ·  \n
  ·  · · ·  \n

Sample Output 3

Copy
3 11
 ·  \n

Hints

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.

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