Bottles CodeForces

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.

Example

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.

题解：dp[ i ][ j ][ k ]表示1~i个瓶子中已选j个瓶子得到的最大实际装水量。那么dp[ i ][ j ][ k ]=max(dp[ i ] [ j ] [ k ]，dp[ i - 1 ][ j - 1][ k - b[ i ] ] + a[ i ] )。

首先，题目要求最少的瓶子最短的时间，而最少的瓶子满足其总容量大于->实际水的总量<-（sumN）。所需要的时间，则是水的总量（sumN）减去这k个瓶子装的水的总量。所以time=sum1-maxSum(K)。

dp好难，初始化都不好掌握 ，><

#include<bits/stdc++.h>
using namespace std;

const int maxn=105;

int n,sum1,sum2;
int dp[maxn][maxn*maxn];

struct node{
    int a,b;
    bool operator<(const node& i)const{
        return i.b<b;
    }
}p[maxn];

void Inite(){
    memset(dp,-1,sizeof(dp));
    dp[0][0]=0;
}

void Solve(){
    int tem=0,q=1;
    for(int i=1;i<=n;i++){
        tem+=p[i].b;
        q=i;
        if(tem>=sum1) break;
    }
    for(int i=1;i<=n;i++)
        for(int j=q;j>=1;j--)
            for(int k=sum2;k>=p[i].b;k--) 
                if(~dp[j-1][k-p[i].b]) dp[j][k]=max(dp[j][k],dp[j-1][k-p[i].b]+p[i].a);
    int ans=0;
    for(int i=sum1;i<=sum2;i++) ans=max(ans,dp[q][i]);
    cout<<q<<" "<<sum1-ans<<endl;
}

int main()
{    
    cin>>n;
    Inite();
    sum1=sum2=0;
    for(int i=1;i<=n;i++){ cin>>p[i].a; sum1+=p[i].a; }
    for(int i=1;i<=n;i++){ cin>>p[i].b; sum2+=p[i].b; }
    sort(p+1,p+n+1);
    Solve();
}