#include <iostream>
#include <vector>
#include <deque>
using namespace std;

#define rep(i,n) for(i = 0;i < n;++i)
#define all(v) v.begin(), v.end()
using ll = long long;

int main()
{
    ll i,j;
    ll n, W;
    cin >> n >> W;
    vector<ll> v(n);
    vector<ll> w(n);
    vector<ll> m(n);
    for(i = 0;i < n;++i){
        cin >> v.at(i) >> w.at(i) >> m.at(i);
    }
    vector<vector<ll>> dp(n+1, vector<ll>(W+1,0));
    for(i = 1;i <= n;++i){
        ll a;
        rep(a, w.at(i-1)){
            deque<pair<ll,ll>> deq;
            for(j = 0;j*w.at(i-1) + a <= W;++j){
                ll tmpw = j*w.at(i-1) + a;
                ll val = dp.at(i-1).at(tmpw) - j*v.at(i-1);
                while((!deq.empty()) && deq.back().first <= val) deq.pop_back();
                deq.emplace_back(val, j);
                dp.at(i).at(tmpw) = deq.front().first + j*v.at(i-1);
                if(j - deq.front().second == m.at(i-1)) deq.pop_front();
            }
        }
    }
    cout << dp.at(n).at(W) << endl;
    return 0;
}

### Prompt

Please provide a cpp coded solution 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 <iostream>
#include <vector>
#include <deque>
using namespace std;

#define rep(i,n) for(i = 0;i < n;++i)
#define all(v) v.begin(), v.end()
using ll = long long;

int main()
{
    ll i,j;
    ll n, W;
    cin >> n >> W;
    vector<ll> v(n);
    vector<ll> w(n);
    vector<ll> m(n);
    for(i = 0;i < n;++i){
        cin >> v.at(i) >> w.at(i) >> m.at(i);
    }
    vector<vector<ll>> dp(n+1, vector<ll>(W+1,0));
    for(i = 1;i <= n;++i){
        ll a;
        rep(a, w.at(i-1)){
            deque<pair<ll,ll>> deq;
            for(j = 0;j*w.at(i-1) + a <= W;++j){
                ll tmpw = j*w.at(i-1) + a;
                ll val = dp.at(i-1).at(tmpw) - j*v.at(i-1);
                while((!deq.empty()) && deq.back().first <= val) deq.pop_back();
                deq.emplace_back(val, j);
                dp.at(i).at(tmpw) = deq.front().first + j*v.at(i-1);
                if(j - deq.front().second == m.at(i-1)) deq.pop_front();
            }
        }
    }
    cout << dp.at(n).at(W) << endl;
    return 0;
}

```