CodeForces 366C Dima and Salad



Dima, Inna and Seryozha have gathered in a room. That's right, someone's got to go. To cheer Seryozha up and inspire him to have a walk, Inna decided to cook something.

Dima and Seryozha have n fruits in the fridge. Each fruit has two parameters: the taste and the number of calories. Inna decided to make a fruit salad, so she wants to take some fruits from
the fridge for it. Inna follows a certain principle as she chooses the fruits: the total taste to the total calories ratio of the chosen fruits must equal
k. In other words,

, where
aj is the taste of the
j-th chosen fruit and
bj is its calories.

Inna hasn't chosen the fruits yet, she is thinking: what is the maximum taste of the chosen fruits if she strictly follows her principle? Help Inna solve this culinary problem — now the happiness of a young couple is in your
hands!

Inna loves Dima very much so she wants to make the salad from at least one fruit.

Input
The first line of the input contains two integers
n, k
(1 ≤ n ≤ 100, 1 ≤ k ≤ 10). The second line of the input contains
n integers a1, a2, ..., an
(1 ≤ ai ≤ 100) — the fruits' tastes. The third line of the input contains
n integers b1, b2, ..., bn
(1 ≤ bi ≤ 100) — the fruits' calories. Fruit number
i has taste ai and calories
bi.

Output
If there is no way Inna can choose the fruits for the salad, print in the single line number -1. Otherwise, print a single integer — the maximum possible sum of the taste values of the chosen fruits.

Examples

Input

3 2
10 8 1
2 7 1

Output

18

Input

5 3
4 4 4 4 4
2 2 2 2 2

Output

-1

Note
In the first test sample we can get the total taste of the fruits equal to 18 if we choose fruit number 1 and fruit number 2, then the total calories will equal 9. The condition


fulfills, that's exactly what Inna wants.

In the second test sample we cannot choose the fruits so as to follow Inna's principle.

题意:n个物品，k为倍数。每个物品有两个属性（ai和bi），求在满足所取物品的a属性和是b属性和的k倍的前提下，问a属性的最大值是多少，

题解：

背包，不过得先处理两个属性，由于最后要求a属性是b属性的k倍，所以我们把每个物品的ai-bi*k看作重量，把ai看作价值，进行背包，不过ai-bi*k有可能为负数，这里有两种处理方法，1：将重量同时增大，使其都变为正数。2；正数的做一次背包，负数的做一次背包，求和。

代码：

#include<cstring>
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<fstream>
#include <algorithm>
#include <cmath>
#define M 20000
using namespace std;
int dp[M],pp[M];
int a[M],b[M],c[M];
int n,k;
int main()
{
while(~scanf("%d%d",&n,&k))
{
memset(dp,-0x3f ,sizeof(dp));
memset(pp,-0x3f ,sizeof(pp));
dp[0]=pp[0]=0;
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(int i=0;i<n;i++)
{
scanf("%d",&b[i]);
c[i]=a[i]-b[i]*k;
}
for(int i=0;i<n;i++)
{
if(c[i]>=0)
{
for(int j=10000;j>=c[i];j--)
{
dp[j]=max(dp[j],dp[j-c[i]]+a[i]);
}
}
else
{
c[i]=-c[i];
for(int j=10000;j>=c[i];j--)
{
pp[j]=max(pp[j],pp[j-c[i]]+a[i]);
}

}
}
int ans=-1;
for(int i=0;i<=10000;i++)
{
if(dp[i]==0&&pp[i]==0)
continue;
else
ans=max(dp[i]+pp[i],ans);
}
cout<<ans<<endl;
}
}