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 REP(i, n) for(int i = 0;i < n;i++)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end())
#define VRSORT(v) sort(v.rbegin(), v.rend())
#define ll long long
#define pb(a) push_back(a)
#define INF 1000000000
#define LINF 1e18
#define MOD 1000000007
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;

typedef vector<unsigned int>vec;
typedef vector<vec> mat;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline void add(T &a, T b){a = ((a+b) % MOD + MOD) % MOD;};

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const double PI = acos(-1.0);
const double EPS = 1e-10;

#define MAX_N 40
ll w[MAX_N],v[MAX_N];
pair<ll,ll> ps[1<<(MAX_N/2)];

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N;
    ll W;
    cin>>N>>W;
    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];
            }
        }
        ps[i] = make_pair(sw,sv);
    }

    sort(ps,ps+(1<<n2));
    ll m=1;
    for(int i=1;i<(1<<n2);i++){
        if(ps[m-1].second<ps[i].second) ps[m++] = ps[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){
            ll tv = (lower_bound(ps,ps+m,LP(W-sw,LINF))-1)->second;
            chmax(res,sv+tv);
        }
    }
    cout<<res<<endl;
}

