Dima and Salad(完全背包)

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 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 integers a₁, 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 numberi 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.

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.

题解：

做水果色拉，有n种水果，水果有两个属性：味道和卡路里，选一定水果，让求使得味道和除以卡路里和为k的时候的味道和的最大值；

刚开始我的想法是弄一个完全背包，可以得到多种组合；判断bag[j] / j == k即可；但是当味道和一定的时候可能存在多个卡路里和值，必然会对下面的数据造成影响；

看了大神的，是转化了下：c =a -b*k;可以转化成两个背包；正数存一个，负数存一个，以免造成当味道和一定的时候可能存在多个卡路里和值的情况；

ac

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
    class Program
    {
        static void getdata(ref int x) {
            x = ;
            int t;
            while (true) {
                t = Console.Read();
                if (t >= '' && t <= '') break;
            }
            while (true) {
                if (t < '' || t > '') break;
                x = x *  + t - '';
                t = Console.Read();
            }
        }
        static int max(int a, int b) {
            return a > b ? a : b;
        }
        static int min(int a, int b)
        {
            return a < b ? a : b;
        }
        static void Main(string[] args)
        {
            int[] a = new int[];
            int[] b = new int[];
            int[] bag1 = new int[];
            int[] bag2 = new int[];
            int n = , k = ;
            getdata(ref n);
            getdata(ref k);
            //Console.Read();
            for (int i = ; i < ; i++) {
                bag1[i] = -;
                bag2[i] = -;
            }
            bag1[] = ;
            bag2[] = ;
            int ans = , V = ;
                for (int i = ; i < n; i++)
                {
                    getdata(ref a[i]);
                }
            for (int i = ; i < n; i++) {
                getdata(ref b[i]);
            }

            for (int i = ; i < n; i++) {
                int w = a[i] - b[i] * k;
                if (w < )
                {
                    w = -w;
                    for (int j = V; j >= w; j--)
                    {
                        bag1[j] = max(bag1[j], bag1[j - w] + a[i]);
                    }
                }
                else {
                    for (int j = V; j >= w; j--) {
                        bag2[j] = max(bag2[j], bag2[j - w] + a[i]);
                    }
                }
            }
            for (int i = ; i >= ;i-- )
            {
                ans = max(ans, bag1[i] + bag2[i]);
            }
                if (ans == )
                    Console.WriteLine(-);
                else
                    Console.WriteLine(ans);
              // while (true) ;
        }
    }
}