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++)
#define all(obj) (obj).begin(),(obj).end()
typedef long long ll;
#define mod 1e9+7
const ll inf= 1e17;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N,W;
    cin>>N>>W;
    vector<pair<ll,ll> > p;
    ll v[40],w[40];
    rep(i,N)cin>>v[i]>>w[i];
    int n2=N/2;
    for(int i=0;i<1<<n2;i++){
        ll sw=0,sv=0;
        for(int j=0;j<n2;j++){
            if(i>>j&1){
                sw+=w[j];
                sv+=v[j];
            }
        }
        p.push_back(make_pair(sw,sv));
    }
    sort(all(p));
    int m=1;
    for(int i=1;i<1<<n2;i++)if(p[m-1].second<p[i].second) p[m++]=p[i];
    ll res=0;
     for(int i=0;i<1<<(N-n2);i++){
        ll sw=0,sv=0;
        for(int j=0;j<(N-n2);j++){
            if(i>>j&1){
                sw+=w[n2+j];
                sv+=v[n2+j];
            }
        }
        if(sw<=W){
            pair<ll,ll> pp=make_pair(W-sw,inf);
            ll tv=(lower_bound(p.begin(),p.begin()+m,pp)-1)->second;
            res=max(res,sv+tv);
        }
    }
    cout<<res<<endl;
    return 0;
}

