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;

int main()
{
	ll N, W;
	cin >> N >> W;
	vector<pair<ll, ll>> vw(N);
	for (int i = 0; i < N; i++) {
		cin >> vw[i].first >> vw[i].second;
	}
	int x = N / 2, y = N - N / 2;
	vector<pair<ll, ll>> A(1 << x), B(1 << y);
	for (int i = 0; i < (1 << x); i++) {
		for (int j = 0; j < x; j++) {
			if (i & (1 << j)) {
				A[i].second += vw[j].first;
				A[i].first += vw[j].second;
			}
		}
	}
	sort(A.begin(), A.end());
	for (int i = 1; i < (1 << x); i++) {
		A[i].second = max(A[i].second, A[i - 1].second);
	}
	for (int i = 0; i < (1 << y); i++) {
		for (int j = 0; j < y; j++) {
			if (i & (1 << j)) {
				B[i].second += vw[x + j].first;
				B[i].first += vw[x + j].second;
			}
		}
	}
	ll res = 0;
	for (auto b : B) {
		auto elb = upper_bound(A.begin(), A.end(), pair<ll, ll>(W - b.first, 1000000000000000000ll));
		if(elb != A.begin())
			res = max(res, b.second + (elb - 1)->second);
	}
	cout << res << endl;
	return 0;
}