Buns

​​傳送門​​ Lavrenty, a baker, is going to make several buns with stuffings and sell them.

Lavrenty has n grams of dough as well as m different stuffing types. The stuffing types are numerated from 1 to m. Lavrenty knows that he has ai grams left of the i-th stuffing. It takes exactly bi grams of stuffing i and ci grams of dough to cook a bun with the i-th stuffing. Such bun can be sold for di tugriks.

Also he can make buns without stuffings. Each of such buns requires c0 grams of dough and it can be sold for d0 tugriks. So Lavrenty can cook any number of buns with different stuffings or without it unless he runs out of dough and the stuffings. Lavrenty throws away all excess material left after baking.

Find the maximum number of tugriks Lavrenty can earn.

Input

The first line contains 4 integers n, m, c0 and d0 (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10, 1 ≤ c0, d0 ≤ 100). Each of the following m lines contains 4 integers. The i-th line contains numbers ai, bi, ci and di (1 ≤ ai, bi, ci, di ≤ 100).

Output

思路：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int a[1010],b[1010],c[1010],d[1010],num[1010];
int dp[1010];

int main()
{
  int n,m,c0,d0;
  scanf("%d%d%d%d",&n,&m,&c0,&d0);
  for(int i = 1; i <= m; i++)
  {
    scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
    num[i] = a[i] / b[i];
    a[i] = a[i] % b[i];
  }
  for(int i = c0; i <= n; i++)
  {
    dp[i] = i/c0*d0;
  }
  for(int i = 1; i <= m; i++)
  {
    for(int j = 0; j < num[i]; j++)
    {
      for(int k = n; k >= c[i]; k--)
      {
        dp[k] = max(dp[k],dp[k-c[i]] + d[i]);
      }
    }
  }
  printf("%d\n",dp[n]);
}