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;

using ll = long long;
using P = pair<ll, ll>;

constexpr ll INF = 1e18;

int main() {
    ll N, W;
    cin >> N >> W;
    vector<ll> w(N), v(N);
    for(int i=0; i<N; ++i) {
        cin >> v[i] >> w[i];
    }
    ll n2 = N/2;
    vector<P> ps(1 << n2);
    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.begin(), ps.end());
    int 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.begin(), ps.begin() + m, make_pair(W-sw, INF)) - 1)->second;
            res = max(res, sv + tv);
        }
    }
    cout << res << endl;
}
