Bottles CodeForces - 730J

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

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?=?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 ，><

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3
 4 const int maxn=105;
 5
 6 int n,sum1,sum2;
 7 int dp[maxn][maxn*maxn];
 8
 9 struct node{
10     int a,b;
11     bool operator<(const node& i)const{
12         return i.b<b;
13     }
14 }p[maxn];
15
16 void Inite(){
17     memset(dp,-1,sizeof(dp));
18     dp[0][0]=0;
19 }
20
21 void Solve(){
22     int tem=0,q=1;
23     for(int i=1;i<=n;i++){
24         tem+=p[i].b;
25         q=i;
26         if(tem>=sum1) break;
27     }
28     for(int i=1;i<=n;i++)
29         for(int j=q;j>=1;j--)
30             for(int k=sum2;k>=p[i].b;k--)
31                 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);
32     int ans=0;
33     for(int i=sum1;i<=sum2;i++) ans=max(ans,dp[q][i]);
34     cout<<q<<" "<<sum1-ans<<endl;
35 }
36
37 int main()
38 {
39     cin>>n;
40     Inite();
41     sum1=sum2=0;
42     for(int i=1;i<=n;i++){ cin>>p[i].a; sum1+=p[i].a; }
43     for(int i=1;i<=n;i++){ cin>>p[i].b; sum2+=p[i].b; }
44     sort(p+1,p+n+1);
45     Solve();
46 }