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;

pair<ll, ll> ps[1 << (40 / 2)];

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int N;
    ll W;
    cin >> N >> W;
    ll w[N], v[N];

    for (int i = 0; i < N; ++i) {
        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));

    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, ps + m, make_pair(W - sw, (ll)1e15 )) - 1) ->second;
            res = max(res, sv + tv);
        }
    }

    cout << res << '\n';

    return 0;
}

