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>
#define ll long long
#define INF 999999999
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<n;i++)

using namespace std;

typedef pair<int,int>P;
typedef pair<ll,ll>pl;

const int MAX_N = 40;

ll N,W;
ll v[MAX_N];
ll w[MAX_N];
vector<pl> x,y;

void func(int si,int ti,vector<pl> &x){
  int size=ti-si;
  rep(bit,1<<size){
    pl p=pl(0,0);
    rep(i,size){
      if(bit>>i&1){
        p.first+=w[si+i];
        p.second+=v[si+i];
      }
    }
    x.push_back(p);
  }
}
  
int main(){
  cin>>N>>W;
  rep(i,N){
  	cin>>v[i]>>w[i];
  }
 
  func(0,N/2,x);
  sort(x.begin(),x.end());
  func(N/2,N,y);
  sort(y.begin(),y.end());
    
  ll target=0,maxm=-(1LL<<60),ans=0;
    
  for(int i=x.size()-1;i>=0;i--){
    pl p=x[i];
    while(target<y.size() && p.first+y[target].first <=W ){
      maxm=max(maxm,y[target].second);
      target++;
    }
    ans=max(ans,maxm+p.second);
  }
  
  cout<<ans<<endl;
  return 0;
}