text: # Problem 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.# Solution #include <bits/stdc++.