Problem: 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

Solution (in CPP):

Solution:
/* monkukui 競技プログラミング用のテンプレート  (ここから) */
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <functional>
#include <bitset>

using namespace std;
using lint = long long int;
using ll = long long int;
using lnt = long long int;
using Graph = vector<vector<int>>;
using Wgraph = vector<vector<pair<int, long long int>>>;
constexpr long long int INF = 1001001001001001LL;
constexpr long long int MOD = 1000000007LL;
constexpr double PI = 3.1415926535897932;
constexpr long long int di[] = {-1, 0, 1, 0, -1, 1, 1, -1};
constexpr long long int dj[] = {0, 1, 0, -1, 1, 1, -1, -1};

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
inline void yes(){ cout << "yes" << endl; }
inline void Yes(){ cout << "Yes" << endl; }
inline void YES(){ cout << "YES" << endl; }
inline void no(){ cout << "no" << endl; }
inline void No(){ cout << "No" << endl; }
inline void NO(){ cout << "NO" << endl; }
inline void possible(){ cout << "possible" << endl; }
inline void Possible(){ cout << "Possible" << endl; }
inline void POSSIBLE(){ cout << "POSSIBLE" << endl; }
inline void impossible(){ cout << "impossible" << endl; }
inline void Impossible(){ cout << "Impossible" << endl; }
inline void IMPOSSIBLE(){ cout << "IMPOSSIBLE" << endl; }

#define REP(i,n) for(int i = 0; i < (n); i++)
#define RREP(i,n) for(int i = 1; i <= (n); i++)
#define DREP(i,n) for(int i = (n)-1; i >= 0; i--)
#define SREP(i,s,t) for(int i = s; i < t; i++)
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define PB push_back

/* monkukui 競技プログラミング用のテンプレート (ここまで)*/

vector<long long> KnapsackProblemWithLimitations(vector<long long> v, vector<long long> w, vector<long long> m, long long W) {
  
  size_t n = v.size();
  vector<long long> dp(W + 1, 0);
  for(size_t i = 0; i < n; i++) {
    for(long long k = 0; m[i] > 0; k++) {
      long long key = min(m[i], (long long)(1 << k));
        m[i] -= key;
        for(long long j = W; j >= key * w[i]; j--) {
        dp[j] = max(dp[j], dp[j - key * w[i]] + key * v[i]);
      }
    }
  }

  return dp;
}

int main(){
  
  lnt n, W; cin >> n >> W;
  vector<lnt> v(n);
  vector<lnt> w(n);
  vector<lnt> m(n);

  for(int i = 0; i < n; i++) cin >> v[i] >> w[i] >> m[i];
  auto dp = KnapsackProblemWithLimitations(v, w, m, W);
  lnt ans = 0;
  for(auto x : dp) ans = max(ans, x);
  cout << ans << endl;
  return 0;
}