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#include <bits/stdc++.h>
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using namespace std;
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const int N = 1010; // 个数上限
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const int M = 2010; // 体积上限
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int n, m, idx;
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int f[M];
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/*
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Q:为什么是N*12?
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A:本题中v_i<=2000,因为c数组是装的打包后的物品集合，每类物品按二进制思想，2000最多可以打log2(2000)+1个包，即 10.96578+1=12个足够,
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同时，共N类物品，所以最大值是N*12。
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如果题目要求v_i<=INT_MAX,那么就是log2(INT_MAX)=31,开31个足够,因为31是准确的数字，不需要再上取整。
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为保险起见，可以不用计算数组上限，直接N*32搞定！
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*/
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struct Node {
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int w, v;
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} c[N * 12];
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int main() {
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cin >> n >> m;
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// 多重背包的经典二进制打包办法
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for (int i = 1; i <= n; i++) {
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int v, w, s;
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cin >> v >> w >> s;
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for (int j = 1; j <= s; j *= 2) { // 1,2,4,8,16,32,64,128,...打包
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c[++idx] = {j * w, j * v};
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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 * w, s * v};
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}
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// 按01背包跑
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for (int i = 1; i <= idx; i++)
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for (int j = m; j >= c[i].v; j--) // 倒序
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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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