#include <iostream>
#include <vector>
using namespace std;
/*
把多重背包问题转化为01背包问题
这样，所有物品就被分为只能用1次和能用无限次
针对vector中物品的不同类别，用对应的状态转移方程计算即可
*/


const int N = 1010;
int n, m;
int v, w, s;
int f[N];
struct Thing
{
    int cate;
    int v, w;
};
std::vector<Thing> things;

int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; ++i)
    {
        cin >> v >> w >> s;
        if(s < 0)
            things.push_back({-1, v, w});
        else if(s == 0)
            things.push_back({0, v, w});
        else
        {
            int k = 1;
            while(k <= s)
            {
                things.push_back({-1, v*k, w*k});
                s -= k;
                k *= 2;
            }
            if(s)
                things.push_back({-1, v*s, w*s});
        }
    }

    for(auto thing : things)
    {
        if(thing.cate < 0)
            for (int j=m; j>=thing.v; j--)
                f[j] = max(f[j], f[j-thing.v]+thing.w);
        else
            for (int j=thing.v; j<=m; j++)
                f[j] = max(f[j], f[j-thing.v]+thing.w);
    }

    cout << f[m];

    return 0;
}