CodeForces - 366C（01背包）

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

3 210 8 1
2 7 1

Output

18

Input

5 34 4 4 4 4
2 2 2 2 2

Output

-1

 

题目大意：

有很多物品，都有a，b两个值，要求选择多个物品，使得 

，求满足上面条件的，这些物品 a 值和的最大值。

思路：

可以把这个式子退一下，变成01背包，但是重量有负值，所以可以整体偏移一个值，再来计算。

代码：

 

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <cmath>
#include <algorithm>
#include <map>
#include <queue>
#include <cmath>
#include <ctime>
using namespace std;
#define LL long long
const int maxn = 110;
const int mx = 100000;
int n, k;
int a[maxn], b[maxn];
int dp[maxn][210000];

int main()
{

  scanf("%d%d", &n, &k);

  for (int i = 1;i <= n;i++)
  {
    scanf("%d", &a[i]);
  }
  for (int i = 1;i <= n;i++)
  {
    scanf("%d", &b[i]);
  }
  memset(dp, -0x3f3f3f, sizeof(dp));
  dp[0][mx] = 0;
  for (int i = 1;i <= n;i++)
  {
    int c = a[i] - k*b[i];
    for (int j = mx*2;j >= 0;j--)
    {
      if (j - c<0 || j - c>mx * 2)continue;
      dp[i][j] = max(dp[i-1][j], dp[i - 1][j - c] + a[i]);
    }
  }
  if (dp[n][mx] == 0)puts("-1");
  else printf("%d\n", dp[n][mx]);
  return 0;
}