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 <iostream>
#include <string>
#include <map>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <queue>
#include <tgmath.h>
#define MOD 1000000007
using namespace std;
#define INT_MAX 2147000000
#define LONG_MAX 9223372036854775806
typedef long long ll;

#define rep(i,n) for(int i=0;i<n;i++)
ll N,W;
#define MAXN 40
ll v[MAXN],w[MAXN];
ll INF = 1<<21;
pair<ll,ll> P[1<<(MAXN/2)];
int main(){
    cin >> N >> W;
    rep(i,N)cin >> v[i] >> w[i];
    int half = N/2;
    for(int i=0;i<1<<half;i++){
        ll tmpw = 0;
        ll tmpv = 0;
        for(int j=0;j<half;j++){
            if(i>>j&1){
                tmpw += w[j];
                tmpv += v[j];
            }
        }
        P[i] = make_pair(tmpw,tmpv);
    }
    ll m = 1;
    sort(P,P+(1<<half));
    for(int i=1;i<1<<half;i++){
        if(P[m-1].second<P[i].second){
            P[m++] = P[i];
        }
    }
    ll res = 0;
    for(int i=0;i<1<<(N-half);i++){
        ll tmpw = 0;
        ll tmpv = 0;
        for(int j=0;j<(N-half);j++){
            if(i>>j&1){
                tmpw += w[j+half];
                tmpv += v[j+half];
            }
        }
        if(W>=tmpw){
            ll tv = (lower_bound(P,P+m,make_pair(W-tmpw,INF))-1)->second;
            res = max(res,tv+tmpv);
        }
    }

    cout << res << endl;
    return 0;
}
