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 sys

input = sys.stdin.buffer.readline

n = int(input())
soda = list(map(int, input().split()))
volume = list(map(int, input().split()))
dp = [[{} for j in range(n + 1)] for k in range(n + 1)]
dp[0][1][volume[0] - soda[0]] = 0
total = soda[0]
for i in range(0, n - 1):
    total += soda[i + 1]
    dp[i + 1][1][volume[i + 1] - total] = total - soda[i + 1]
    for bottle in range(1, i + 2):
        for volume_left in dp[i][bottle]:
            if volume_left - soda[i + 1] not in dp[i + 1][bottle]:
                dp[i + 1][bottle][volume_left - soda[i + 1]] = dp[i][bottle][volume_left] + soda[i + 1]
            dp[i + 1][bottle][volume_left - soda[i + 1]] = min(dp[i][bottle][volume_left] + soda[i + 1],dp[i + 1][bottle][volume_left - soda[i + 1]])
            if volume_left + volume[i + 1] - soda[i + 1] not in dp[i + 1][bottle + 1]:
                dp[i + 1][bottle + 1][volume_left + volume[i + 1] - soda[i + 1]] = dp[i][bottle][volume_left]
            dp[i + 1][bottle + 1][volume_left + volume[i + 1] - soda[i + 1]] = min(dp[i][bottle][volume_left],dp[i + 1][bottle + 1][volume_left + volume[i + 1] -soda[i + 1]])
flag = 0
mini = 99999999999
i = 0
for i in range(1, n + 1):
    for j in dp[n - 1][i]:
        if j >= 0:
            mini = min(mini, dp[n - 1][i][j])
            flag = 1
    if flag == 1:
        break
print(i, mini)