FJ is going to do some shopping, and before that, he needs some boxes to carry the different kinds of stuff he is going to buy. Each box is assigned to carry some specific kinds of stuff (that is to say, if he is going to buy one of these stuff, he has to buy the box beforehand). Each kind of stuff has its own value. Now FJ only has an amount of W dollars for shopping, he intends to get the highest value with the money.

Input

The first line will contain two integers, n(the number of boxes 1 ≤ n ≤ 50), w (the amount of money FJ has, 1 ≤ w ≤ 100000) Then n lines follow. Each line contains the following number p_i (the price of the ith box 1 ≤ p_i ≤ 1000), m_i (1 ≤ m_i ≤ 10 the number goods i-th box can carry), and m_i pairs of numbers, the price c_j (1 ≤ c_j ≤ 100), the value v_j (1 ≤ v_j ≤ 1000000).

Output

For each test case, output the maximum value FJ can get

Sample Input

3 800

300 2 30 50 25 80

600 1 50 130

400 3 40 70 30 40 35 60

Sample Output

210

【题目大意】给若干组物品，每组物品都有一个箱子（箱子自身也有cost），然后就是物品的cost和value，要买某个物品必须也要买装这个物品的箱子，给一定钱数，问能获得的最大价值。

解题思路：对每个箱子的附件进行一次01背包，背包的容量是总花费-箱子的花费，得到的就是对应每个附件的最大价值；然后再对这个箱子进行一次01背包，箱子的价值dpbox[i-p]与不取这个箱子的价值dptotal[i]那个大取哪个，模模糊糊知道大体思路

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 using namespace std;

 const int MAX =  + ;

 int dptotal[MAX],dpbox[MAX];

 int price,value,n,w,p,c,m;

 int main()

 {

     while(scanf("%d%d", &n,&w) != EOF)

     {

         memset(dptotal, , sizeof(dptotal));

         for(int i = ; i < n; i++)

         {

             scanf("%d%d",&p,&m);

             memcpy(dpbox, dptotal, sizeof(dptotal));

             for(int k = ; k <= m; k++)

             {

                 scanf("%d%d",&price,&value);

                 for(int j = w - p; j >= price; j--)

                 {

                     dpbox[j] = max(dpbox[j], dpbox[j - price] + value);

                 }

             }

             for(int k = w; k >= p; k--)

             {

                 if(dptotal[k] < dpbox[k - p])

                     dptotal[k] = dpbox[k - p];

             }

         }

         printf("%d\n",dptotal[w]);

     }

     return ;

 }