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;

constexpr int inf = 0x3f3f3f3f;
constexpr ll linf = 0x3f3f3f3f3f3f3f3fLL;

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n; ll W; cin >> n >> W;
    vector<ll> v(n), w(n);
    for (int i = 0; i < n; i++) cin >> v[i] >> w[i];

    int m = n >> 1;
    vector<pair<ll, ll>> l(1 << m, make_pair(0, 0)), r(1 << (n - m), make_pair(0, 0));
    for (int s = 0; s < (1 << m); s++) {
        for (int i = 0; i < m; i++) {
            if ((s >> i) & 1) l[s].first += w[i], l[s].second += v[i];
        }
    }
    for (int s = 0; s < (1 << (n - m)); s++) {
        for (int i = 0; i < n - m; i++) {
            if ((s >> i) & 1) r[s].first += w[m + i], r[s].second += v[m + i];
        }
    }
    sort(l.begin(), l.end());

    vector<pair<ll, ll>> lu;
    lu.emplace_back(l.front());
    for (int i = 1; i < (1 << m); i++) {
        if (lu.back().second < l[i].second) lu.emplace_back(l[i]);
    }

    ll ans = 0;
    for (auto &p: r) {
        ll wl = p.first, vl = p.second;
        if (wl > W) continue;
        ll vr = (lower_bound(lu.begin(), lu.end(), make_pair(W - wl, linf)) - 1)->second;
        ans = max(ans, vl + vr);
    }
    cout << ans << endl;
    return 0;
}
