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<n;i++)
#define REP(i,s,n) for(int i=s;i<n;i++)
typedef long long ll;
const ll INF = pow(2, 63) - 1;
int main(){
    ll n, W;
    cin >> n >> W;
    vector<pair<ll, ll> > p;
    vector<ll> v(n), w(n);
    rep(i,n)cin >> v[i] >> w[i];
    int n2 = n / 2;
    rep(i,1 << n2){
        ll sv = 0, sw = 0;
        rep(j,n2)if(i >> j & 1){
            sv += v[j];
            sw += w[j];
        }
        p.push_back(make_pair(sw, sv));
    }
    sort(p.begin(), p.end());
    int m = 1;
    REP(i,1,1<<n2)if(p[m-1].second < p[i].second){
        p[m] = p[i];
        m++;
    }
    ll r = 0;
    rep(i, 1 << (n-n2)) {
        ll sw = 0, sv = 0;
        rep(j, n - n2)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 tmp_v = (lower_bound(p.begin(), p.begin() + m, pp) - 1)->second;
            r = max(r, sv + tmp_v);
        }
    }
    cout << r << endl;
    return 0;
}
