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 <algorithm>
#include <map>
#include <vector>

using ll = long long;

int main() {
    int N;
    ll W;
    std::cin >> N >> W;

    std::vector<ll> v(N), w(N);
    for (int i = 0; i < N; ++i) std::cin >> v[i] >> w[i];

    std::vector<std::pair<ll, ll>> vp;
    int M = N / 2;
    for (int b = 0; b < (1 << M); ++b) {
        ll wsum = 0, vsum = 0;
        for (int i = 0; i < M; ++i) {
            if ((b >> i) & 1) {
                wsum += w[i];
                vsum += v[i];
            }
        }
        vp.emplace_back(wsum, vsum);
    }

    // 重さについて価値が昇順になるよう記録
    std::sort(vp.begin(), vp.end());
    std::map<ll, ll> fore;
    ll vmax = -1;
    for (auto& p : vp) {
        if (p.second <= vmax) continue;
        fore[p.first] = p.second;
        vmax = p.second;
    }

    ll ans = 0;
    for (int b = 0; b < (1 << (N - M)); ++b) {
        ll wsum = 0, vsum = 0;
        for (int i = 0; i < N - M; ++i) {
            if ((b >> i) & 1) {
                wsum += w[i + M];
                vsum += v[i + M];
            }
        }
        if (wsum > W) continue;
        auto itr = fore.upper_bound(W - wsum);
        ans = std::max(ans, vsum + (--itr)->second);
    }
    std::cout << ans << std::endl;
    return 0;
}

