CF--Dima and Salad

Dima and Salad

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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 a_j is the taste of the j-th chosen fruit and b_j 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 integersa₁, a₂, ..., a_n(1 ≤ a_i ≤ 100) — the fruits' tastes. The third line of the input contains n integers b₁, b₂, ..., b_n(1 ≤ b_i ≤ 100) — the fruits' calories. Fruit number i has taste a_i and calories b_i.

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.

Sample Input

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

Hint

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.

题目大意:给你n种水果,每一种水果都有两个属性,口味和卡路里,接着要做一道水果沙拉,选出的水果必须满足如下条件,选出的所有水果的口味之和/卡路里之和=k.寻找满足该条件的最大口味之和.

解体思路:对于题目中所给的公式,可以如下转换,(ax+ay+az+...)/(bx+by+bz...)=k;  -->   (ax-k*bx)+(ay-k*by)+(az-k*bz)+...=0;

令c=a[i]-k*b[i];得到的c有正有负，那么我们可以分开dp背包（即将c值进行组合），c值互为相反数，那么就满足题目中的要求，即口味之和/卡路里之和=k。

代码如下：

#include
#include
#include
#define INF 0x3f3f3f3f
using namespace std;
int dp1[10010],dp2[10010];
int a[110],b[110];
int main(){
	int n,k,i,j;
	while(scanf("%d%d",&n,&k)!=EOF){
		for(i=1;i<=n;i++){
			scanf("%d",&a[i]);
		}
		for(i=1;i<=n;i++){
			scanf("%d",&b[i]);
			b[i]=a[i]-k*b[i];
		}
		for(i=1;i<=10000;i++){
			dp1[i]=-INF;
			dp2[i]=-INF;
		}
		dp1[0]=dp2[0]=0;
		for(i=1;i<=n;i++){
		     if(b[i]>=0){
		     	for(j=10000;j>=b[i];j--){
		     		dp1[j]=max(dp1[j],dp1[j-b[i]]+a[i]);
		     	}
		     }
		     else{
		     	b[i]=-b[i];
		     	for(j=10000;j>=b[i];j--){
		     			dp2[j]=max(dp2[j],dp2[j-b[i]]+a[i]);
		     	}
		     }
		}
//		printf("%d %d\n",dp2[6],dp1[6]);
		int ans=-INF;
		for(i=10000;i>=0;i--){
			if(dp1[i]==0&&dp2[i]==0) continue;
			ans=max(ans,dp1[i]+dp2[i]);
		}
		if(ans<0) printf("-1\n");
		else printf("%d\n",ans);

	}
	return 0;
}