Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.

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 $m_i$ items for $i$-th item.



Find the maximum total value of items in the knapsack.

Constraints

* $1 \le N \le 50$
* $1 \le v_i \le 50$
* $1 \le w_i \le 10^9$
* $1 \le m_i \le 10^9$
* $1 \le W \le 10^9$

Input


$N$ $W$
$v_1$ $w_1$ $m_1$
$v_2$ $w_2$ $m_2$
:
$v_N$ $w_N$ $m_N$


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


Input

5 1000000000
3 5 1000000000
7 6 1000000000
4 4 1000000000
6 8 1000000000
2 5 1000000000


Output

1166666666

Solution (in CPP):

Solution:
// includes
#include <bits/stdc++.h>

// macros
#define ll long long int
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define vrep(v, i) for(int i = 0; i < (v).size(); i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define FI first
#define SE second
#define dump(a, n) for(int i = 0; i < n; i++)cout << a[i] << "\n "[i + 1 != n];
#define dump2(a, n, m) for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)cout << a[i][j] << "\n "[j + 1 != m];
#define bit(n) (1LL<<(n))
#define INT(n) int n; cin >> n;
#define LL(n) ll n; cin >> n;
#define DOUBLE(n) double n; cin >> n;
using namespace std;

//  types
typedef pair<int, int> P;
typedef pair<ll, int> Pl;
typedef pair<ll, ll> Pll;
typedef pair<double, double> Pd;
typedef complex<double> cd;
 
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

// solve
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}

const int MAX_N = 50*50*50 + 1;
ll dp[MAX_N];

int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  INT(n); LL(W);
  vector<int> v(n);
  vector<ll> w(n), m(n);
  rep(i, n)cin >> v[i] >> w[i] >> m[i];

  rep(i, MAX_N)dp[i] = linf;
  rep(i, n){
    int k = min(50LL, m[i]);
    m[i] -= k;
    for(int j = 0; j < k; j++){
      dp[0] = 0;
      for(int l = MAX_N-1; l > 0; l--){
        if(l-v[i]>=0)dp[l] = min(dp[l], w[i] + dp[l-v[i]]);
      }
    }
  }

  vector<Pll> vec(n);
  rep(i, n)vec[i] = mk(v[i], w[i]);
  vector<int> c(n);
  rep(i, n)c[i] = i;
  sort(all(c), [&](int i, int j){
      return v[i] * w[j] > v[j] * w[i];
      });
  ll res = 0;
  rep(i, MAX_N){
    ll rem = W - dp[i];
    ll ans = 0;
    rep(j, n){
      if(rem <= 0)break;
      ll r = min(m[c[j]], rem / w[c[j]]);
      rem -= r * w[c[j]];
      ans += r * v[c[j]];
    }
    if(rem >= 0)res = max(res, ans + i);
  }
  cout << res << endl;
  return 0;
}