You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.

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.



Find the maximum total value of items in the knapsack.

Constraints

* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015

Input


N W
v1 w1
v2 w2
:
vN wN


The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.

Output

Print the maximum total values of the items in a line.

Examples

Input

4 5
4 2
5 2
2 1
8 3


Output

13


Input

2 20
5 9
4 10


Output

9
#include <iostream>
#include <vector>
#include <algorithm>

#define int long long
#define rep(i,n) for(int i=0;i<(n);i++) 

using namespace std;
using P = pair<int, int>;

const int inf = 1e18;
int N,W;
int v[40],w[40];
vector<P> ps;
vector<P> qs;

signed main(void) {
  cin >> N >> W;
  rep(i,N) {
    cin >> v[i] >> w[i];
  }

  rep(i,(1<<N/2)) {
    int sw = 0, sv = 0;
    rep (j, N/2) {
      if ((i & (1ll << j)) != 0) {
        sw += w[j];
        sv -= v[j];
      }
    }
    if (sw <= W) ps.push_back(P(sw, sv));
  }

  sort(ps.begin(), ps.end());
  for(auto &v : ps) v.second *= -1;

  for (auto v : ps) {
    if (qs.empty() || qs.back().second < v.second) {
      qs.push_back(v);
    }
  }

  int ans = 0;
  rep(i,(1<<(N-N/2))) {
    int sw = 0, sv = 0;
    for (int j = N / 2; j < N; j++) {
      if ((i & (1ll << (j - N / 2))) != 0) {
        sw += w[j];
        sv += v[j];
      }
    }
    if (sw <= W) {
      ans = max(ans, (lower_bound(qs.begin(), qs.end(), P(W - sw, inf)) - 1)->second + sv);
    }
  }
  cout << ans << endl;
  return 0;
}
