2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest - J. Bottles(DP)

Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volumebi (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 xseconds 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

43 3 4 34 7 6 5

output

2 6

input

21 1100 100

output

1 1

input

510 30 5 6 2410 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 = 6seconds to do it.

分析：f[i][j] 表示当前用了i个瓶子，共堆出来j的容量的最大剩余.

#include<iostream>#include<string>#include<algorithm>#include<cstdlib>#include<cstdio>#include<set>#include<map>#include<vector>#include<cstring>#include<stack>#include<queue>#define INF 4557430888798830400ll#define eps 1e-9#define MAXN 0x3f#define N 105using namespace std;int n,k,ans,tot,tot_b,cnt,a[N],s[N],b[N],f[N][10005]; int main(){scanf("%d",&n);for(int i = 1;i <= n;i++) {scanf("%d",&a[i]);tot += a[i];}for(int i = 1;i <= n;i++) {scanf("%d",&b[i]);s[i] = b[i];tot_b += b[i];}sort(s+1,s+1+n);for(int i = 1;i <= n && cnt < tot;i++){k++;cnt += s[n-i+1];}memset(f,-1,sizeof(f));f[0][0] = 0,ans = tot;for(int i = 1;i <= n;i++) for(int j = tot_b;j >= b[i];j--)  for(int l = k;l;l--)   if(f[l-1][j-b[i]] >= 0) f[l][j] = max(f[l][j],f[l-1][j-b[i]] + a[i]);for(int j = tot_b;j >= tot;j--) ans = min(ans,tot - f[k][j]);printf("%d %d\n",k,ans);    }