C. Dima and Salad
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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
Copy 
3 2
10 8 1
2 7 1
output 
18
input
Copy 
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：w[i]=a[i]-k*b[i],v[i]=a[i]，问选择若干个，使得在背包容量为0的情况下，获得最大的v。由于w[i]可正，可负。 分开2次DP。

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

const int MAX_N=1e6+3;
const int MOD=1e9+7;
const int M=1e6;
int dp1[MAX_N];
int dp2[MAX_N];
int w[MAX_N],v[MAX_N];
int n,m;

int main(void){
    int n,k;
    cin >> n>>k;
    for(int i=1;i<=n;++i)   scanf("%d",&v[i]);
    for(int i=1;i<=n;++i)   scanf("%d",&w[i]);
    memset(dp2,-0x3f,sizeof(dp2));
    memset(dp1,-0x3f,sizeof(dp1));
    dp1[0]=dp2[0]=0;
    for(int i=1;i<=n;i++){
        int ans=v[i]-k*w[i];
        if(ans>=0){
            for(int j=M;j>=ans;j--)
                dp1[j]=max(dp1[j],dp1[j-ans]+v[i]);
        }
        else{
            ans=-ans;
            for(int j=M;j>=ans;j--)
                dp2[j]=max(dp2[j],dp2[j-ans]+v[i]);
        }
    }

    int res=-1;
    for(int i=0;i<=M;i++)
        res=max(res,dp1[i]+dp2[i]);//两个等大反值的背包
    if(res==0)  cout << "-1" <<endl;
    else
    cout <<res << endl;
}