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
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
 
#include <vector>
#include <set>
#include <cstdio>
using namespace std;
  
int main(){
    int N;
    long long W,tV=0,tW=0,x=0;
    scanf("%d%lld",&N,&W);
    vector<pair<long long,long long> >v(N);
    for(int i=0;i<N;i++)scanf("%lld%lld",&v[i].first,&v[i].second),tV+=v[i].first,tW+=v[i].second;
 
        int n=N/2;
        set<pair<long long,long long> >se1,se2;
        for(int i=0;i<1<<n;i++){
            long long V=0,w=0;
            for(int j=0;j<n;j++)if(i&(1<<j))V+=v[j].first,w+=v[j].second;
            se1.emplace(w,V);
        }
        for(int i=0;i<1<<(N-n);i++){
            long long V=0,w=0;
            for(int j=0;j<(N-n);j++)if(i&(1<<j))V+=v[j+n].first,w+=v[j+n].second;
            se2.emplace(w,V);
        }
        for(auto it=se1.begin();it!=se1.end();){
            auto it2=it;
            for(++it2;it2!=se1.end()&&it->second>=it2->second;)it2=se1.erase(it2);
            ++it;
        }
        for(auto it=se2.begin();it!=se2.end();){
            auto it2=it;
            for(++it2;it2!=se2.end()&&it->second>=it2->second;)it2=se2.erase(it2);
            ++it;
        }
        long long r=0;
        for(auto &e1:se1){
            auto it=se2.lower_bound({W-e1.first+1,0});
            if(it!=se2.begin()){
                --it;
                r=max(r,e1.second+it->second);
            }
        }
        printf("%lld\n",r);
}