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#include <bits/stdc++.h>
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using namespace std;
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const int N = 1e5 + 10;
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int n; // 物品种类
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int m; // 背包容量
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int f[N]; // dp数组
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int idx;
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struct Node {
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int v, w;
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} c[N * 12];
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int main() {
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cin >> n >> m;
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// 二进制打包
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for (int i = 1; i <= n; i++) {
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// 体积,价值,个数
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int v, w, s;
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cin >> v >> w >> s;
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// 根据题意做一些小的变形
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if (s == -1)
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s = 1; // 题目中s=-1表示只有1个
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else if (s == 0)
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s = m / v; // 完全背包(其实本质上就是多重背包):最多总体积/该物品体积向下取整
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// 如果是其它大于0的数字,那么是多重背包
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// 将完全背包和多重背包利用二进制优化转化为01背包
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for (int j = 1; j <= s; j *= 2) {
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c[++idx] = {j * v, j * w};
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s -= j;
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}
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// 不够下一个2^n时,独立成包
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if (s) c[++idx] = {s * v, s * w};
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}
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// 01背包
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for (int i = 1; i <= idx; i++)
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for (int j = m; j >= c[i].v; j--)
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f[j] = max(f[j], f[j - c[i].v] + c[i].w);
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// 输出
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printf("%d\n", f[m]);
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return 0;
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}
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