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

J. Bottles

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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 a₁, a₂, ..., 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 b₁, b₂, ..., 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.

/*
背包问题：最少用几个桶很简单能求出来。然后背包出这些桶最少
*/

/*
比赛的时候怎么就脑残的想到了贪心没想到背包啊！！！赛后出题真是啧啧啧
*/
#include <bits/stdc++.h>
using namespace std;
struct qwe
{
    int a,b;
} e[110];
int cmp(qwe x,qwe y)
{
    return (x.b>y.b || (x.b==y.b && x.a>y.a));//体积大的排序，如果体积相同就按照已经有水多的的在前面
}
int cmp1(qwe x,qwe y)//按照体积小的排序
{
    return (x.b<y.b);
}
int dp[110][10010],s[110];
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    int n;
    scanf("%d",&n);
    int sum=0;//总共有多少水
    for(int i=1;i<=n;i++) 
        scanf("%d",&e[i].a),sum+=e[i].a;
    for(int i=1;i<=n;i++) 
        scanf("%d",&e[i].b);
    sort(e+1,e+1+n,cmp);
    int t=0,q=0;
    int ans;
    for(int i=1;i<=n;i++)
    {
        t+=e[i].b;
        if(t>=sum)
        {
            ans=i;
            break;
        }
    }//找出最少的桶数
    sort(e+1,e+1+n,cmp1);
    for(int i=1;i<=n;i++) 
        s[i]=s[i-1]+e[i].b;
    memset(dp,0xBf,sizeof(dp));
    dp[0][0]=0;//dp[j][k]表示j个桶，最多装多少水
    for(int i=1;i<=n;i++)//用多少桶
    {
        for(int j=min(i,ans);j>0;j--)
            for (int k=s[i];k>=e[i].b;k--)
            {
                    dp[j][k]=max(dp[j][k],dp[j-1][k-e[i].b]+e[i].a);
            }
    }
    int fin=0;
    for (int k=s[n];k>=sum;k--)
        fin=max(fin,dp[ans][k]);
    printf("%d %d\n",ans,sum-fin);
    return 0;
}