#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const ll MOD = 1000000007;
const int IINF = INT_MAX;
const ll LLINF = LLONG_MAX;
const int MAX_N = int(1e5 + 5);
const double EPS = 1e-10;
const int di[] = {0, 1, 0, -1}, dj[] = {1, 0, -1, 0};
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define SORT(v) sort((v).begin(), (v).end())
#define ALL(v) (v).begin(), (v).end()

int N, W, M, dp[2000][10005];
vector<int> v, w;

int main() {
    cin >> N >> W;
    REP(i,N){
        int vi, wi, m, b=1;
        cin >> vi >> wi >> m;
        // divide m into 1, 2, 4, 8...
        while(b<=m){
            v.push_back(vi*b);
            w.push_back(wi*b);
            m -= b;
            b *= 2;
        }
        // the rest of m
        if(m>0){
            v.push_back(vi*m);
            w.push_back(wi*m);
        }
    }
    M = v.size();
    REP(i,M){
        REP(j,W+1){
            dp[i+1][j] = max(dp[i+1][j], dp[i][j]);
            if(j >= w[i]) dp[i+1][j] = max(dp[i+1][j], dp[i][j-w[i]]+v[i]);
        }
    }
    cout << dp[M][W] << endl;
    return 0;
}


### Prompt

Construct a CPP code solution to the problem outlined:

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>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const ll MOD = 1000000007;
const int IINF = INT_MAX;
const ll LLINF = LLONG_MAX;
const int MAX_N = int(1e5 + 5);
const double EPS = 1e-10;
const int di[] = {0, 1, 0, -1}, dj[] = {1, 0, -1, 0};
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define SORT(v) sort((v).begin(), (v).end())
#define ALL(v) (v).begin(), (v).end()

int N, W, M, dp[2000][10005];
vector<int> v, w;

int main() {
    cin >> N >> W;
    REP(i,N){
        int vi, wi, m, b=1;
        cin >> vi >> wi >> m;
        // divide m into 1, 2, 4, 8...
        while(b<=m){
            v.push_back(vi*b);
            w.push_back(wi*b);
            m -= b;
            b *= 2;
        }
        // the rest of m
        if(m>0){
            v.push_back(vi*m);
            w.push_back(wi*m);
        }
    }
    M = v.size();
    REP(i,M){
        REP(j,W+1){
            dp[i+1][j] = max(dp[i+1][j], dp[i][j]);
            if(j >= w[i]) dp[i+1][j] = max(dp[i+1][j], dp[i][j-w[i]]+v[i]);
        }
    }
    cout << dp[M][W] << endl;
    return 0;
}


```