//朴素解法
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 110;

int dp[N][N], v[N], w[N], s[N];

int main()
{
    int n, V;
    cin >> n >> V;
    for(int i = 1; i <= n; i++) cin >> v[i] >> w[i] >> s[i];

    for(int i = 1; i <= n; i++)
    {
        for(int j = 0; j <= V; j++)
        {
            for(int k = 0; k <= s[i] && k * v[i] <= j; k++)
            {
                dp[i][j] = max(dp[i][j], dp[i - 1][j - k * v[i]] + k * w[i]);
            }
        }
    }

    cout << dp[n][V] << endl;

    return 0;
}

//优化解法      （二进制拆分优化）
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 25000, M = 2000;

int v[N], w[N], dp[N];

int main()
{
    int n, V;
    cin >> n >> V;
    int cnt = 0;
    for(int i = 1; i <= n; i++)
    {
        int a, b, s;
        cin >> a >> b >> s;
        int k = 1;
        while(k <= s)
        {
            cnt++;
            v[cnt] = a * k;
            w[cnt] = b * k;
            s = s - k;
            k = k * 2;
        }
        if(s > 0)
        {
            cnt++;
            v[cnt] = a * s;
            w[cnt] = b * s;
        }
    }

    n = cnt;

    for(int i = 1; i <= n; i++)
    {
        for(int j = V; j >= v[i]; j--)
        {
            dp[j] = max(dp[j], dp[j - v[i]] + w[i]);
        }
    }

    cout << dp[V] << endl;

    return 0;
}