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 <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
using P = pair<ll,ll>;
const ll INF = 1e18;

int main(){
  ll n, W;
  cin >> n >> W;
  ll v[n], w[n];
  rep(i,n) cin >> v[i] >> w[i];
  vector<P> dp;
  rep(i,1<<(n/2)) {
    ll val = 0, wei = 0;
    rep(j,n/2) {
      if(i & (1<<j)) {
        val += v[j];
        wei += w[j];
      }
    }
    dp.push_back(P(wei,val));
  }
  sort(dp.begin(),dp.end());
  rep(i,dp.size()) {
    if(i > 0) dp[i].second = max(dp[i].second,dp[i-1].second);
  }
  ll ans = 0;
  rep(i,1<<(n-n/2)) {
    ll val = 0, wei = 0;
    rep(j,n-n/2) {
      if(i & (1<<j)) {
        val += v[n/2+j];
        wei += w[n/2+j];
      }
    }
    if(wei <= W) {
      int id = upper_bound(dp.begin(),dp.end(),P(W-wei,INF)) - dp.begin();
      id--;
      if(id >= 0) val += dp[id].second;
      ans = max(ans,val);
    } 
  }
  cout << ans << endl;
  return 0;
}
