#include <bits/stdc++.h>

#define M_PI       3.14159265358979323846   // pi

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> VI;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> t3;
typedef tuple<ll, ll, ll, ll> t4;

#define rep(a,n) for(ll a = 0;a < n;a++)
#define repi(a,b,n) for(ll a = b;a < n;a++)

static const ll INF = 1e15;

const ll mod = 1000000007;

vector<ll> knapsack_limitation(const vector<ll> & w, const vector<ll> & m, const vector<ll> & v, const ll weight) {
	const int inv = -1;
	const int n = w.size();
	vector<ll> dp(weight + 1, inv);
	dp[0] = 0;
	rep(i, n) {
		for (int a = 0; a < w[i]; a++) {
			deque<ll> ids, vals;
			for (int j = 0; w[i] * j + a <= weight; j++) {
				ll cw = w[i] * j + a;
				if (cw != inv) {
					ll val = dp[cw] - j * v[i];
					while (!vals.empty() && val > vals.back()) {
						ids.pop_back();
						vals.pop_back();
					}
					ids.push_back(j);
					vals.push_back(val);
				}
				if (ids.size()) {
					dp[cw] = vals.front() + j * v[i];
					if (ids.front() == j - m[i]) {
						ids.pop_front();
						vals.pop_front();
					}
				}
			}
		}
	}
	return dp;
}


int main(void)
{
	int N, W;
	cin >> N >> W;
	vector<ll> v(N), w(N), m(N);
	for (int i = 0; i < N; i++) {
		cin >> v[i] >> w[i] >> m[i];
	}
	auto ret = knapsack_limitation(w, m, v, W);
	cout << *max_element(begin(ret), end(ret)) << endl;
}


### Prompt

Develop a solution in cpp to the problem described below:

You have N items that you want to put them into a knapsack. Item i has value vi, weight wi and limitation mi.

You want to find a subset of items to put such that:

* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
* You can select at most mi items for ith item.



Find the maximum total value of items in the knapsack.

Constraints

* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 1,000
* 1 ≤ wi ≤ 1,000
* 1 ≤ mi ≤ 10,000
* 1 ≤ W ≤ 10,000

Input


N W
v1 w1 m1
v2 w2 m2
:
vN wN mN


The first line consists of the integers N and W. In the following N lines, the value, weight and limitation of the i-th item are given.

Output

Print the maximum total values of the items in a line.

Examples

Input

4 8
4 3 2
2 1 1
1 2 4
3 2 2


Output

12


Input

2 100
1 1 100
2 1 50


Output

150

### Response

```cpp
#include <bits/stdc++.h>

#define M_PI       3.14159265358979323846   // pi

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> VI;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> t3;
typedef tuple<ll, ll, ll, ll> t4;

#define rep(a,n) for(ll a = 0;a < n;a++)
#define repi(a,b,n) for(ll a = b;a < n;a++)

static const ll INF = 1e15;

const ll mod = 1000000007;

vector<ll> knapsack_limitation(const vector<ll> & w, const vector<ll> & m, const vector<ll> & v, const ll weight) {
	const int inv = -1;
	const int n = w.size();
	vector<ll> dp(weight + 1, inv);
	dp[0] = 0;
	rep(i, n) {
		for (int a = 0; a < w[i]; a++) {
			deque<ll> ids, vals;
			for (int j = 0; w[i] * j + a <= weight; j++) {
				ll cw = w[i] * j + a;
				if (cw != inv) {
					ll val = dp[cw] - j * v[i];
					while (!vals.empty() && val > vals.back()) {
						ids.pop_back();
						vals.pop_back();
					}
					ids.push_back(j);
					vals.push_back(val);
				}
				if (ids.size()) {
					dp[cw] = vals.front() + j * v[i];
					if (ids.front() == j - m[i]) {
						ids.pop_front();
						vals.pop_front();
					}
				}
			}
		}
	}
	return dp;
}


int main(void)
{
	int N, W;
	cin >> N >> W;
	vector<ll> v(N), w(N), m(N);
	for (int i = 0; i < N; i++) {
		cin >> v[i] >> w[i] >> m[i];
	}
	auto ret = knapsack_limitation(w, m, v, W);
	cout << *max_element(begin(ret), end(ret)) << endl;
}


```