Nick 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.

Difficulty: competition

Starter code:

Solutions:
f = lambda: list(map(int, input().split()))
n = int(input())
a, b = f(), f()

d = [[None] * 10001 for i in range(n)]

def g(i, s):
    if s <= 0: return (0, s)
    if i == n: return (1e7, 0)

    if not d[i][s]:
        x, y = g(i + 1, s - b[i])
        d[i][s] = min(g(i + 1, s), (x + 1, y + b[i] - a[i]))
    return d[i][s]

x, y = g(0, sum(a))
print(x, y)