https://codeforces.com/problemset/problem/730/J

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?aiis 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.

C++版本一

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUGusing namespace std; typedef long long ll; const int N=100+10; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m; struct node {int a,b;bool operator < (const node & S)const{if(b!=S.b)return b>S.b;elsereturn a>S.a;} }e[N]; int dp[N][N*N]; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endifscanf("%d",&n);int sum=0;memset(dp,-1,sizeof(dp));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+n+1);int tmp=0;int cnt=0,ans=0;for(int i=1;i<=n;i++){tmp+=e[i].b;if(tmp>=sum){cnt=i;break;}}dp[0][0]=0;for(int i=1;i<=n;i++){for(int j=sum;j>=e[i].a;j--){for(int k=i;k>=1;k--){if(dp[k-1][j-e[i].a]!=-1)dp[k][j]=max(dp[k][j],dp[k-1][j-e[i].a]+e[i].b);}}}for(int i=sum;i>=1;i--){if(dp[cnt][i]>=sum){ans=sum-i;break;}}cout << cnt <<" "<< ans << endl;//cout << "Hello world!" << endl;return 0; }

C++版本二

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