// http://tsutaj.hatenablog.com/entry/2017/01/25/161050
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define P pair<ll,ll>
#define FOR(I,A,B) for(ll I = (A); I < (B); ++I)
#define FORR(I,A,B) for(ll I = (B-1); I >= (A); --I)
#define SORT(x) (sort(x.begin(),x.end())) // 0 2 2 3 4 5 8 9
#define REV(x) (reverse(x.begin(),x.end())) //reverse
ll gcd(ll a,ll b){if(a<b)swap(a,b);if(a%b==0)return b;else return gcd(b,a%b);}
ll lcm(ll a,ll b){ll c=gcd(a,b);return ((a/c)*(b/c)*c);}//saisyo kobaisu
const ll INF=1e18+7;
const ll MOD=1e9+7;


struct KnapWithLimitations{
	ll N,W;
	vector<ll> v,w,m,dp;
	KnapWithLimitations(ll n,ll ww){
		N=n;W=ww;
		v.resize(N);
		w.resize(N);
		m.resize(N);
		dp.resize(W+1,0);
	}
	ll maxvalue(){
		FOR(i,0,N){
			for(ll k=0;m[i]>0;k++){
				ll key = min(m[i],(ll)(1<<k));
				m[i] -= key;
				for(ll j=W;j>=key*w[i];j--){
					dp[j] = max(dp[j],dp[j-key*w[i]]+key*v[i]);
				}
			}
		}
		ll ans = 0;
		FOR(i,0,W+1)ans=max(ans,dp[i]);
		return ans;
	}
};



int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	ll n,w;
	cin >> n >> w;
	KnapWithLimitations KL(n,w);
	FOR(i,0,n){
		cin >> KL.v[i] >> KL.w[i] >> KL.m[i];
	}
	cout << KL.maxvalue() << endl;
}


### Prompt

Please create a solution in cpp to the following 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

### Response

```cpp
// http://tsutaj.hatenablog.com/entry/2017/01/25/161050
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define P pair<ll,ll>
#define FOR(I,A,B) for(ll I = (A); I < (B); ++I)
#define FORR(I,A,B) for(ll I = (B-1); I >= (A); --I)
#define SORT(x) (sort(x.begin(),x.end())) // 0 2 2 3 4 5 8 9
#define REV(x) (reverse(x.begin(),x.end())) //reverse
ll gcd(ll a,ll b){if(a<b)swap(a,b);if(a%b==0)return b;else return gcd(b,a%b);}
ll lcm(ll a,ll b){ll c=gcd(a,b);return ((a/c)*(b/c)*c);}//saisyo kobaisu
const ll INF=1e18+7;
const ll MOD=1e9+7;


struct KnapWithLimitations{
	ll N,W;
	vector<ll> v,w,m,dp;
	KnapWithLimitations(ll n,ll ww){
		N=n;W=ww;
		v.resize(N);
		w.resize(N);
		m.resize(N);
		dp.resize(W+1,0);
	}
	ll maxvalue(){
		FOR(i,0,N){
			for(ll k=0;m[i]>0;k++){
				ll key = min(m[i],(ll)(1<<k));
				m[i] -= key;
				for(ll j=W;j>=key*w[i];j--){
					dp[j] = max(dp[j],dp[j-key*w[i]]+key*v[i]);
				}
			}
		}
		ll ans = 0;
		FOR(i,0,W+1)ans=max(ans,dp[i]);
		return ans;
	}
};



int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	ll n,w;
	cin >> n >> w;
	KnapWithLimitations KL(n,w);
	FOR(i,0,n){
		cin >> KL.v[i] >> KL.w[i] >> KL.m[i];
	}
	cout << KL.maxvalue() << endl;
}


```