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J. Bottles

time limit per test

2 secondsmemory limit per test

512 megabytesinput

standard inputoutput

standard outputNick has *n* bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda *a*_{i} and bottle volume *b*_{i} (*a*_{i} ≤ *b*_{i}).

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 *a*_{1}, *a*_{2}, ..., *a*_{n} (1 ≤ *a*_{i} ≤ 100), where *a*_{i} is the amount of soda remaining in the *i*-th bottle.

The third line contains *n* positive integers *b*_{1}, *b*_{2}, ..., *b*_{n} (1 ≤ *b*_{i} ≤ 100), where *b*_{i} is the volume of the *i*-th bottle.

It is guaranteed that *a*_{i} ≤ *b*_{i} 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.

Codeforces (c) Copyright 2010-2017 Mike Mirzayanov

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